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\paragraph{Question 2(a)}  Show that the sequence $x(n)=\frac{1}{n}+\frac{1}{n+1}+\ldots +\frac{1}{2n}$ is monotone decreasing and bounded below and therefore convergent.   x(n) is monotone decreasing if $x(n+1)\leq x(n)$ for all $n\geq 1$.    \[ x(n+1) = \frac{1}{n+1}+\frac{1}{n+2}+\ldots +\frac{1}{2n+1} \]    Since \[ \frac{1}{n} > \frac{1}{n+1} \] for $n \geq 1$     \\x(n) naturally starts off greater. The denominator of the fraction being added increases for each x(n) and x(n+1) so the sequence converges.    \\ $\therefore x(n)=\frac{1}{n}+\frac{1}{n+1}+\ldots +\frac{1}{2n} > x(n+1) = \frac{1}{n+1}+\frac{1}{n+2}+\ldots +\frac{1}{2n+1}$ and is monotone decreasing.     \\ The sequence x(n) is bounded below if there is a real number L such that $x(n)\geq L$ for all $n\geq 1$.  Lets say L = 2 then $x(n) \geq 1$ for all $n\geq 1$.  \\When n = 1.  \[\frac{1}{1} = 1+..+\]  Then we add positive rational numbers that are less than 1 so the sum is less than 2.  \\As n increases, n=1 is greater than any other n because $\frac{1}{n}$ the n is in the denominator.  \\ $\therefore$ there is a real number L such that $x(n)\geq L$ for all $n\geq 1$.    \paragraph{(b)} Calculate and plot x(n) for $1\leq n \leq 1000$ and estimate $\lim_{n\to \infty}x(n)$.