Let:

\[\begin{aligned} S=\sum^{\infty}_{n=1} n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)} \end{aligned}\]

If I use the Ratio Test to determine whether \(S\) converges, I need to determine:

\[\begin{aligned} \lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right| \end{aligned}\]

What is the value of this limit? \[\begin{aligned} L\:&=\:\lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2(n\:+\:1)\:+\:1)}\hspace{1em}\div\hspace{1em}n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:2\:+\:1)}\hspace{1em}\div\hspace{1em}n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3)}\hspace{1em}\div\hspace{1em}n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3)}\hspace{1em}\cdot\hspace{1em}\frac{1}{n\cdot\left(\frac{5}{4}\right)^{(2n\:+\:1)}}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3)}\cdot\left(\frac{5}{4}\right)^{-(2n\:+\:1)}\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3)}\cdot\left(\frac{5}{4}\right)^{(-2n\:-\:1)}\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{(2n\:+\:3\:-\:2n\:-\:1)}\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{5}{4}\right)^{2}\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:(n\:+\:1)\cdot\left(\frac{25}{16}\right)\hspace{1em}\cdot\hspace{1em}\frac{1}{n}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{25(n\:+\:1)}{16n}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{25n\left(\frac{n}{n}\:+\:\frac{1}{n}\right)}{16n}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{25\left(1\:+\:\frac{1}{n}\right)}{16}\:\right| \\[1em] \:&=\:\left|\:\frac{25\left(1\:+\:\frac{1}{\infty}\right)}{16}\:\right| \\[1em] \:&=\:\left|\:\frac{25\left(1\:+\:0\right)}{16}\:\right| \\[1em] \:&=\:\left|\:\frac{25\left(1\right)}{16}\:\right| \\[1em] \:&=\:\left|\:\frac{25}{16}\:\right| \\[1em] L\:&=\:\left[\frac{25}{16}\right] \end{aligned}\] If \(L < 1\) : \(S\) Converges
If \(L > 1\) or \(\infty\): \(S\) Diverges
If \(L = 1\): \(S\) Inconclusive
Using the above answer, we know that \(S\)
\(L=\frac{25}{16} > 1\) : \(S\) Diverges

Let: \[\begin{aligned} S=\sum^{\infty}_{n=0} \frac{9^n}{(n+1)\cdot2^{(2n-1)}}\end{aligned}\]

If I use the Ratio Test to determine whether \(S\) converges, I need to determine:

\[\begin{aligned} \lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right| \end{aligned}\]

What is the value of this limit?

\[\begin{aligned} L\:&=\:\lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9^\left(n\:+\:1\right)}{(n\:+\:1\:+\:1)\cdot2^{(2(n\:+\:1)\:-\:1)}}\hspace{1em}\div\hspace{1em}\frac{9^n}{(n\:+\:1)\cdot2^{(2n\:-\:1)}}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9^\left(n\:+\:1\right)}{(n\:+\:1\:+\:1)\cdot2^{(2n\:+\:2\:-\:1)}}\hspace{1em}\div\hspace{1em}\frac{9^n}{(n\:+\:1)\cdot2^{(2n\:-\:1)}}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9^\left(n\:+\:1\right)}{(n\:+\:2)\cdot2^{(2n\:+\:1)}}\hspace{1em}\div\hspace{1em}\frac{9^n}{(n\:+\:1)\cdot2^{(2n\:-\:1)}}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9^\left(n\:+\:1\right)}{(n\:+\:2)\cdot2^{(2n\:+\:1)}}\hspace{1em}\cdot\hspace{1em}\frac{(n\:+\:1)\cdot2^{(2n\:-\:1)}}{9^n}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9^\left(n\:+\:1\right)\cdot9^{(-n)}}{(n\:+\:2)\cdot2^{(2n\:+\:1)}}\hspace{1em}\cdot\hspace{1em}\frac{(n\:+\:1)\cdot2^{(2n\:-\:1)}}{1}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9^{\left(n\:+\:1\:-\:n\right)}}{(n\:+\:2)\cdot2^{(2n\:+\:1)}}\hspace{1em}\cdot\hspace{1em}\frac{(n\:+\:1)\cdot2^{(2n\:-\:1)}}{1}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9^{(1)}}{(n\:+\:2)\cdot2^{(2n\:+\:1)}}\hspace{1em}\cdot\hspace{1em}\frac{(n\:+\:1)\cdot2^{(2n\:-\:1)}}{1}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9}{(n\:+\:2)\cdot2^{(2n\:+\:1)}}\hspace{1em}\cdot\hspace{1em}\frac{(n\:+\:1)\cdot2^{(2n\:-\:1)}}{1}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9}{(n\:+\:2)}\hspace{1em}\cdot\hspace{1em}\frac{(n\:+\:1)\cdot2^{(2n\:-\:1)}\cdot2^{-(2n\:+\:1)}}{1}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9}{(n\:+\:2)}\hspace{1em}\cdot\hspace{1em}\frac{(n\:+\:1)\cdot2^{(2n\:-\:1)}\cdot2^{(-2n\:-\:1)}}{1}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9}{(n\:+\:2)}\hspace{1em}\cdot\hspace{1em}\frac{(n\:+\:1)\cdot2^{(2n\:-\:1\:-\:2n\:-\:1)}}{1}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9}{(n\:+\:2)}\hspace{1em}\cdot\hspace{1em}\frac{(n\:+\:1)\cdot2^{(-2)}}{1}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9}{(n\:+\:2)}\hspace{1em}\cdot\hspace{1em}\frac{(n\:+\:1)}{2^{2}}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9}{(n\:+\:2)}\hspace{1em}\cdot\hspace{1em}\frac{(n\:+\:1)}{4}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9(n\:+\:1)}{4(n\:+\:2)}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9n\left(\frac{n}{n}\:+\:\frac{1}{n}\right)}{4n\left(\frac{n}{n}\:+\:\frac{2}{n}\right)}\:\right| \\[1em] \:&=\:\lim_{n\to\infty}\left|\:\frac{9\left(1\:+\:\frac{1}{n}\right)}{4\left(1\:+\:\frac{2}{n}\right)}\:\right| \\[1em] \:&=\:\left|\:\frac{9\left(1\:+\:\frac{1}{\infty}\right)}{4\left(1\:+\:\frac{2}{\infty}\right)}\:\right| \\[1em] \:&=\:\left|\:\frac{9\left(1\:+\:0\right)}{4\left(1\:+\:0\right)}\:\right| \\[1em] \:&=\:\left|\:\frac{9\left(1\right)}{4\left(1\right)}\:\right| \\[1em] \:&=\:\left|\:\frac{9}{4}\:\right| \\[1em] L\:&=\:\left[\:\frac{9}{4}\:\right] \\[1em] \end{aligned}\]

If \(L < 1\) : \(S\) Converges
If \(L > 1\) or \(\infty\): \(S\) Diverges
If \(L = 1\): \(S\) Inconclusive
Using the above answer, we know that \(S\)
\(L=\frac{9}{4} > 1\) : \(S\) Diverges
Using the Ratio Test, determine the largest integer value of \(x\) for which the following series converges:

\[\begin{aligned} \sum^{\infty}_{n=1} \frac{(n\:+\:1)\cdot x^n}{2\cdot3^{n}}\end{aligned}\]

\[\begin{aligned} \lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:1\:+\:1)\cdot x^{(n\:+\:1)}}{2\cdot3^{(n\:+\:1)}}\hspace{1em}\div\hspace{1em}\frac{(n\:+\:1)\cdot x^n}{2\cdot3^{n}}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x^{(n\:+\:1)}}{2\cdot3^{(n\:+\:1)}}\hspace{1em}\div\hspace{1em}\frac{(n\:+\:1)\cdot x^n}{2\cdot3^{n}}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x^{(n\:+\:1)}}{2\cdot3^{(n\:+\:1)}}\hspace{1em}\cdot\hspace{1em}\frac{2\cdot3^{n}}{(n\:+\:1)\cdot x^n}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x^{(n\:+\:1)}\cdot x^{(-n)}}{2\cdot3^{(n\:+\:1)}}\hspace{1em}\cdot\hspace{1em}\frac{2\cdot3^{n}}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x^{(n\:+\:1\:-\:n)}}{2\cdot3^{(n\:+\:1)}}\hspace{1em}\cdot\hspace{1em}\frac{2\cdot3^{n}}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x^{(1)}}{2\cdot3^{(n\:+\:1)}}\hspace{1em}\cdot\hspace{1em}\frac{2\cdot3^{n}}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x}{2\cdot3^{(n\:+\:1)}}\hspace{1em}\cdot\hspace{1em}\frac{2\cdot3^{n}}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x}{2}\hspace{1em}\cdot\hspace{1em}\frac{2\cdot3^{n}\cdot3^{-(n\:+\:1)}}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x}{2}\hspace{1em}\cdot\hspace{1em}\frac{2\cdot3^{n}\cdot3^{(-n\:-\:1)}}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x}{2}\hspace{1em}\cdot\hspace{1em}\frac{2\cdot3^{(n\:-\:n\:-\:1)}}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x}{2}\hspace{1em}\cdot\hspace{1em}\frac{2\cdot3^{(-1)}}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x}{2}\hspace{1em}\cdot\hspace{1em}\frac{2}{3(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x}{2}\hspace{1em}\cdot\hspace{1em}\frac{2\cdot3}{3(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x}{2}\hspace{1em}\cdot\hspace{1em}\frac{2}{3(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{2x(n\:+\:2)}{2\cdot3(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x(n\:+\:2)}{3(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x\cdot n\left(\frac{n}{n}\:+\:\frac{2}{n}\right)}{3n\left(\frac{n}{n}\:+\:\frac{1}{n}\right)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \left|\:\frac{x \left(1\:+\:\frac{2}{\infty}\right)}{3\left(1\:+\:\frac{1}{\infty}\right)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \left|\:\frac{x \left(1\:+\:0\right)}{3\left(1\:+\:0\right)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \left|\:\frac{x \left(1\right)}{3\left(1\right)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \left|\:\frac{x}{3}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \left|\:x\:\right|\hspace{1em}<\hspace{1em}3 \\[1em]\end{aligned}\]

\( MaxInt \hspace{1em}=\hspace{1em}3\:-\:1\hspace{1em}=\hspace{1em}2\)

What is the radius of convergence of the series:

\[\begin{aligned} \sum^{\infty}_{n=0} \frac{(n\:+\:1)\cdot x^{(n\:+\:1)}}{(n\:+\:3)}\end{aligned}\]

Apply Ratio Test: \[\begin{aligned} \lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:1\:+\:1)\cdot x^{(n\:+\:1\:+\:1)}}{(n\:+\:3\:+\:1)}\hspace{1em}\div\hspace{1em}\frac{(n\:+\:1)\cdot x^{(n\:+\:1)}}{(n\:+\:3)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x^{(n\:+\:2)}}{(n\:+\:4)}\hspace{1em}\div\hspace{1em}\frac{(n\:+\:1)\cdot x^{(n\:+\:1)}}{(n\:+\:3)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x^{(n\:+\:2)}}{(n\:+\:4)}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:3)}{(n\:+\:1)\cdot x^{(n\:+\:1)}}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x^{(n\:+\:2)}\cdot x^{-(n\:+\:1)}}{(n\:+\:4)}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:3)}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x^{(n\:+\:2)}\cdot x^{(-n\:-\:1)}}{(n\:+\:4)}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:3)}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x^{(n\:+\:2\:-\:n\:-\:1)}}{(n\:+\:4)}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:3)}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x^{(1)}}{(n\:+\:4)}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:3)}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x}{(n\:+\:4)}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:3)}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:2)\cdot x}{(n\:+\:4)}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:3)}{(n\:+\:1)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:3)(n\:+\:2)\cdot x}{(n\:+\:1)(n\:+\:4)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n^2\:+\:2n\:+\:3n\:+\:6)\cdot x}{(n^2\:+\:4n\:+\:n\:+\:4)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n^2\:+\:5n\:+\:6)\cdot x}{(n^2\:+\:5n\:+\:4)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{n^2\left(\frac{n^2}{n^2}\:+\:\frac{5n}{n^2}\:+\:\frac{6}{n^2}\right)\cdot x}{n^2\left(\frac{n^2}{n^2}\:+\:\frac{5n}{n^2}\:+\:\frac{4}{n^2}\right)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \lim_{n\to\infty}\left|\:\frac{\left(1\:+\:\frac{5}{n}\:+\:\frac{6}{n^2}\right)\cdot x}{\left(1\:+\:\frac{5}{n}\:+\:\frac{4}{n^2}\right)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \left|\:\frac{\left(1\:+\:\frac{5}{\infty}\:+\:\frac{6}{\infty}\right)\cdot x}{\left(1\:+\:\frac{5}{\infty}\:+\:\frac{4}{\infty}\right)}\:\right|\hspace{1em}<\hspace{1em}1 \\[1em] \left|\:\frac{\left(1\:+\:0\:+\:0\right)\cdot x}{\left(1\:+\:0\:+\:0\right)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{\left(1\right)\cdot x}{\left(1\right)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\: x \:\right|\hspace{1em}<\hspace{1em} 1 \\[1em]\end{aligned}\]

What is the radius of convergence for the series: \[\begin{aligned} \sum^{\infty}_{n=0} \frac{x^n}{(n\:+\:1)!}\end{aligned}\]

Apply Ratio Test: \[\begin{aligned} \lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x^{(n\:+\:1)}}{(n\:+\:1\:+\:1)!}\hspace{1em}\div\hspace{1em}\frac{x^n}{(n\:+\:1)!}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x^{(n\:+\:1)}}{(n\:+\:2)!}\hspace{1em}\div\hspace{1em}\frac{x^n}{(n\:+\:1)!}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x^{(n\:+\:1)}}{(n\:+\:2)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:1)!}{x^n}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x^{(n\:+\:1)}\cdot x^{(-n)}}{(n\:+\:2)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:1)!}{1}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x^{(n\:+\:1\:-\:n)}}{(n\:+\:2)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:1)!}{1}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x^{(1)}}{(n\:+\:2)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:1)!}{1}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x}{(n\:+\:2)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:1)!}{1}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:1)! \cdot x}{(n\:+\:2)!}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(n\:+\:1)! \cdot x}{(n\:+\:2)(n\:+\:1)!}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x}{(n\:+\:2)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x}{n\left(\frac{n}{n}\:+\:\frac{2}{n}\right)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x}{n\left(1\:+\:\frac{2}{n}\right)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{x}{\infty\left(1\:+\:\frac{2}{\infty}\right)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{x}{\infty\left(1\:+\:0\right)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{x}{\infty\left(1\right)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{x}{\infty}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:x\:\right|\hspace{1em}<\hspace{1em} \infty \\[1em]\end{aligned}\]

The series converges for all real numbers so it has an infinite radius of convergence.

What is the radius of convergence of the series:

\[\begin{aligned} \sum^{\infty}_{n=0} \frac{(-1)^{(n\:-\:1)}\cdot (x\:+\:2)^n}{(n\:+\:2)!}\end{aligned}\]

Apply Ratio Test: \[\begin{aligned} \lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(-1)^{(n\:-\:1\:+\:1)}\cdot (x\:+\:2)^{(n\:+\:1)}}{(n\:+\:2\:+\:1)!}\hspace{1em}\div\hspace{1em}\frac{(-1)^{(n\:-\:1)}\cdot (x\:+\:2)^n}{(n\:+\:2)!}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(-1)^{(n)}\cdot (x\:+\:2)^{(n\:+\:1)}}{(n\:+\:3)!}\hspace{1em}\div\hspace{1em}\frac{(-1)^{(n\:-\:1)}\cdot (x\:+\:2)^n}{(n\:+\:2)!}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(-1)^{(n)}\cdot (x\:+\:2)^{(n\:+\:1)}}{(n\:+\:3)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:2)!}{(-1)^{(n\:-\:1)}\cdot (x\:+\:2)^n}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(-1)^{(n)}\cdot (x\:+\:2)^{(n\:+\:1\:-\:n)}}{(n\:+\:3)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:2)!}{(-1)^{(n\:-\:1)}}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(-1)^{(n)}\cdot (x\:+\:2)^{(1)}}{(n\:+\:3)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:2)!}{(-1)^{(n\:-\:1)}}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(-1)^{(n)}\cdot (x\:+\:2)}{(n\:+\:3)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:2)!}{(-1)^{(n\:-\:1)}}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(-1)^{(n)}\cdot (-1)^{-(n\:-\:1)}\cdot (x\:+\:2)}{(n\:+\:3)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:2)!}{1}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(-1)^{(n)}\cdot (-1)^{(-n\:+\:1)}\cdot (x\:+\:2)}{(n\:+\:3)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:2)!}{1}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(-1)^{(n\:-\:n\:+\:1)}\cdot (x\:+\:2)}{(n\:+\:3)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:2)!}{1}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(-1)^{(1)}\cdot (x\:+\:2)}{(n\:+\:3)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:2)!}{1}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{-1\cdot (x\:+\:2)}{(n\:+\:3)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:2)!}{1}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{-1\cdot (x\:+\:2) \cdot (n\:+\:2)!}{(n\:+\:3)!}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{-1\cdot (x\:+\:2) \cdot (n\:+\:2)!}{(n\:+\:3)(n\:+\:2)!}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{-1\cdot (x\:+\:2) \cdot }{(n\:+\:3)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{-1\cdot (x\:+\:2) \cdot }{n\left(\frac{n}{n}\:+\:\frac{3}{n}\right)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{-1\cdot (x\:+\:2) \cdot }{n\left(1\:+\:\frac{3}{n}\right)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{-1\cdot (x\:+\:2)}{\infty\left(1\:+\:\frac{3}{\infty}\right)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{-1\cdot (x\:+\:2)}{\infty\left(1\:+\:0\right)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{-1\cdot (x\:+\:2)}{\infty\left(1\right)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{-1\cdot (x\:+\:2)}{\infty}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:-1\cdot \left(\frac{x}{\infty}\:+\:\frac{2}{\infty}\right)\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:-1\cdot \left(\frac{x}{\infty}\:+\:0\right)\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:-1\cdot \left(\frac{x}{\infty}\right)\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{-x}{\infty}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \frac{x}{\infty} \hspace{1em}<\hspace{1em} 1 \\[1em] x \hspace{1em}<\hspace{1em} \infty \\[1em]\end{aligned}\]

The series converges for all real numbers so it has an infinite radius of convergence.

Determine the radius of convergence of:

\[\begin{aligned} \sum^{\infty}_{n=0} \frac{(x\:-\:4)^n}{5^n}\end{aligned}\]

Apply Ratio Test: \[\begin{aligned} \lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(x\:-\:4)^{(n\:+\:1)}}{5^{(n\:+\:1)}}\hspace{1em}\div\hspace{1em}\frac{(x\:-\:4)^n}{5^n}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(x\:-\:4)^{(n\:+\:1)}}{5^{(n\:+\:1)}}\hspace{1em}\cdot \hspace{1em}\frac{5^n}{(x\:-\:4)^n}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(x\:-\:4)^{(n\:+\:1)}}{5^{(n\:+\:1)}}\hspace{1em}\cdot \hspace{1em}\frac{5^n}{(x\:-\:4)^n}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(x\:-\:4)^{(n\:+\:1)}}{1}\hspace{1em}\cdot \hspace{1em}\frac{5^n \cdot 5^{-(n\:+\:1)}}{(x\:-\:4)^n}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(x\:-\:4)^{(n\:+\:1)}}{1}\hspace{1em}\cdot \hspace{1em}\frac{5^n \cdot 5^{(-n\:-\:1)}}{(x\:-\:4)^n}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(x\:-\:4)^{(n\:+\:1)}}{1}\hspace{1em}\cdot \hspace{1em}\frac{5^{(n\:-\:n\:-\:1)}}{(x\:-\:4)^n}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(x\:-\:4)^{(n\:+\:1)}}{1}\hspace{1em}\cdot \hspace{1em}\frac{5^{(-1)}}{(x\:-\:4)^n}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(x\:-\:4)^{(n\:+\:1)}}{1}\hspace{1em}\cdot \hspace{1em}\frac{1}{5(x\:-\:4)^n}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(x\:-\:4)^{(n\:+\:1)}\cdot (x\:-\:4)^{(-n)}}{1}\hspace{1em}\cdot \hspace{1em}\frac{1}{5}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{(x\:-\:4)^{(n\:+\:1\:-\:n)}}{1}\hspace{1em}\cdot \hspace{1em}\frac{1}{5}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{(x\:-\:4)^{(1)}}{1}\hspace{1em}\cdot \hspace{1em}\frac{1}{5}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{(x\:-\:4)}{1}\hspace{1em}\cdot \hspace{1em}\frac{1}{5}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{(x\:-\:4)}{5}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:(x\:-\:4)\:\right|\hspace{1em}<\hspace{1em} 5 \\[1em] IoC\: : -5\hspace{1em}<\hspace{1em}x\:-\:4\hspace{1em}<\hspace{1em} 5 \\[1em] IoC\: : -5\:+\:4\hspace{1em}<\hspace{1em}x\hspace{1em}<\hspace{1em} 5\:+\:4 \\[1em] IoC\: : -1\hspace{1em}<\hspace{1em}x\hspace{1em}<\hspace{1em} 9 \\[1em] RoC\:=\:\frac{IoC(max)\:-\:IoC(min)}{2} \\[1em] RoC\:=\:\frac{9\:-\:-1}{2} \\[1em] RoC\:=\:\frac{9\:+\:1}{2} \\[1em] RoC\:=\:\frac{10}{2} \\[1em] RoC\:=\:5 \\\end{aligned}\]

Let: \[\begin{aligned} S=\sum^{\infty}_{n=0} \frac{(n\:-\:2)}{(n\:+\:1)!}\end{aligned}\]

If I use the Ratio Test to determine whether S converges, I need to determine:

\[\begin{aligned} \lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right| \\[1em]\end{aligned}\]

Apply Ratio Test:

\[\begin{aligned} L\hspace{1em}&=\hspace{1em}\lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\lim_{n\to\infty}\left|\:\frac{(n\:-\:2\:+\:1)}{(n\:+\:1\:+\:1)!}\hspace{1em}\div\hspace{1em}\frac{(n\:-\:2)}{(n\:+\:1)!}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\lim_{n\to\infty}\left|\:\frac{(n\:-\:1)}{(n\:+\:2)!}\hspace{1em}\div\hspace{1em}\frac{(n\:-\:2)}{(n\:+\:1)!}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\lim_{n\to\infty}\left|\:\frac{(n\:-\:1)}{(n\:+\:2)!}\hspace{1em}\cdot \hspace{1em}\frac{(n\:+\:1)!}{(n\:-\:2)}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\lim_{n\to\infty}\left|\:\frac{(n\:-\:1)(n\:+\:1)!}{(n\:-\:2)(n\:+\:2)!}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\lim_{n\to\infty}\left|\:\frac{(n\:-\:1)(n\:+\:1)!}{(n\:-\:2)(n\:+\:2)(n\:+\:1)!}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\lim_{n\to\infty}\left|\:\frac{(n\:-\:1)}{(n\:-\:2)(n\:+\:2)}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\lim_{n\to\infty}\left|\:\frac{(n\:-\:1)}{(n^2\:+\:2n\:-\:2n\:-\:4)}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\lim_{n\to\infty}\left|\:\frac{(n\:-\:1)}{(n^2\:-\:4)}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\lim_{n\to\infty}\left|\:\frac{n\left(\frac{n}{n}\:-\:\frac{1}{n}\right)}{n^2\left(\frac{n^2}{n^2}\:-\:\frac{4}{n^2}\right)}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\lim_{n\to\infty}\left|\:\frac{\left(1\:-\:\frac{1}{n}\right)}{n\left(1\:-\:\frac{4}{n^2}\right)}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\left|\:\frac{\left(1\:-\:\frac{1}{\infty}\right)}{\infty\left(1\:-\:\frac{4}{\infty}\right)}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\left|\:\frac{\left(1\:-\:0\right)}{\infty\left(1\:-\:0\right)}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\left|\:\frac{\left(1\right)}{\infty\left(1\right)}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\left|\:\frac{1}{\infty}\:\right| \\[1em] \hspace{1em}&=\hspace{1em}\left|\:0\:\right| \\[1em] L\hspace{1em}&=\hspace{1em}0 \\[1em]\end{aligned}\]

If \(L < 1\) : \(S\) Converges
If \(L > 1\) or \(\infty\): \(S\) Diverges
If \(L = 1\): \(S\) Inconclusive
Using the above answer, we know that \(S\)
\(L=0 < 1\) : \(S\) Converges

What is the radius of convergence of the series: \[\begin{aligned} \sum^{\infty}_{n=1} \frac{x^n\cdot \ln(n)}{\ln(n)!}\end{aligned}\]

Apply Ratio Test: \[\begin{aligned} \lim_{n\to\infty}\left|\:a(n\:+\:1)\hspace{1em}\div\hspace{1em}a(n)\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x^{(n\:+\:1)}\cdot \ln(n\:+\:1)}{\ln(n\:+\:1)!}\hspace{1em}\div\hspace{1em}\frac{x^n\cdot \ln(n)}{\ln(n)!}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x^{(n\:+\:1)}\cdot \ln(n\:+\:1)}{\ln(n\:+\:1)!}\hspace{1em}\cdot \hspace{1em}\frac{\ln(n)!}{x^n\cdot \ln(n)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x^{(-n)}\cdot x^{(n\:+\:1)}\cdot \ln(n\:+\:1)}{\ln(n\:+\:1)!}\hspace{1em}\cdot \hspace{1em}\frac{\ln(n)!}{\ln(n)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x^{(-n\:+\:n\:+\:1)}\cdot \ln(n\:+\:1)}{\ln(n\:+\:1)!}\hspace{1em}\cdot \hspace{1em}\frac{\ln(n)!}{\ln(n)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x^{(1)}\cdot \ln(n\:+\:1)}{\ln(n\:+\:1)!}\hspace{1em}\cdot \hspace{1em}\frac{\ln(n)!}{\ln(n)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \lim_{n\to\infty}\left|\:\frac{x\cdot \ln(n\:+\:1)}{\ln(n\:+\:1)!}\hspace{1em}\cdot \hspace{1em}\frac{\ln(n)!}{\ln(n)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em] \left|\:\frac{x\cdot \ln(\infty)}{\ln(\infty)!}\hspace{1em}\cdot \hspace{1em}\frac{\ln(\infty)!}{\ln(\infty)}\:\right|\hspace{1em}<\hspace{1em} 1 \\[1em]\end{aligned}\]