Is this a correct way to show convergence for $\sum_{k=1}^{\infty} x^k (arctan(\frac{1}{k}))^2$? Is this a correct way to show convergence for $\sum_{k=1}^{\infty} x^k (arctan(\frac{1}{k}))^2$?
Ratio test with $lim_{k->\infty} |\frac{a_{k+1}}{a_k}|: \sum_{n=0}^\infty (\frac {\frac{(\frac{1}{k+1})^{2n+1}}{2n+1}}{\frac{(\frac{1}{k})^{2n+1}}{2n+1}})^2=...=\sum_{n=0}^{\infty}(1-\frac{1}{k+1})^{4n+2}=\sum_{n=0}^{\infty}((1-\frac{1}{k+1})^n)^4(1-\frac{1}{k+1})^2$
According to the binomial theorem we have: $\lim_{n->\infty} \lim_{k->\infty}(1-\frac{1}{k+1})^n=\lim_{n->\infty} \lim_{k->\infty} 1-\frac{1}{k+1}+(-\frac{1}{k+1})^2)+...+(-\frac{1}{k+1})^n=\\lim_{n->\infty} 1+0+0^2+...+0^n=1$
So $R=1$ which means that it converges for $|x|<1$.
For $x=1$ we have exactly $\sum_{k=1}^{\infty} (arctan(\frac{1}{k}))^2$, which converges $\implies$ convergence for $x=-1$ also. So convergence $\forall x: |x|\leq 1$.
 A: Set up your limit, then apply L'Hopital's rule.
\begin{align*}
\lim_{k \rightarrow \infty} \left| \frac{a_{k+1}}{a_k} \right| 
    &= \lim_{k \rightarrow \infty} \left| \frac{x^{k+1} \arctan^2 \left( \frac{1}{k+1} \right)}{x^{k} \arctan^2 \left( \frac{1}{k} \right)} \right|  \\
    &= \lim_{k \rightarrow \infty} \left|x  \frac{ \arctan^2 \left( \frac{1}{k+1} \right)}{ \arctan^2 \left( \frac{1}{k} \right)} \right|  \\
    &= \lim_{k \rightarrow \infty} |x|  \frac{ \arctan^2 \left( \frac{1}{k+1} \right)}{ \arctan^2 \left( \frac{1}{k} \right)}  &  &  \text{squares are automatically nonnegative}  \\
    &= |x| \lim_{k \rightarrow \infty} \frac{ \arctan^2 \left( \frac{1}{k+1} \right)}{ \arctan^2 \left( \frac{1}{k} \right)}  \\
    &\overset{L'H}= |x| \lim_{k \rightarrow \infty} \frac{ \frac{1}{1 + \left( \frac{1}{k+1} \right)^2 }}{ \frac{1}{1 + \left( \frac{1}{k} \right)^2 }}  \\
    &= |x| \cdot 1  \\
    &= |x|  \text{.}
\end{align*}
Now the ratio test gives you a condition on the value of this limit to ensure convergence: $|x| < 1$.
Now to the endpoints.  Suppose $x = 1$.  Then we consider $\sum_{k=1}^\infty \arctan^2 \frac{1}{k}$.  One way to proceed is to note $\arctan x < x$ for $x > 0$.  So
$$\sum_{k=1}^\infty \arctan^2 \frac{1}{k} < \sum_{k=1}^\infty \frac{1}{k^2}  \text{,}  $$
which is a convergent series.
Now suppose $x = -1$.  From the previous, we know the sum we obtain is absolutely convergent, hence convergent.  Therefore, the sum converges for $x \in [-1,1]$.
