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\paragraph{Question 1.a} 1.(a)}  Show that the sequence x(n) defined recursively as follows: \[ x(1)=1 \]  \[ x(n)=x(n-1)+ \frac{1}{n^2} \] 
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\\Lets say $U=2$ then $x(n) \leq 2$ for all $n\geq 1$.  \\Since $\frac{1}{n^2}$ is a fraction that is decreasing as n $\rightarrow \infty$ and x(1) = 1 we have a convergence and x(n) is bounded above by 2.   \paragraph{b} \paragraph{(b)}  Calculate and plot x(n) for $1\leq n \leq 1000$ and so get a rough approximation to $\lim_{n\to \infty}x(n)$.