convergence of the series $\sum_{n=1}^{\infty}(1/n^{\alpha})(\int_{0}^{\frac{\pi}{4}}\tan^{n}tdt)x^{n}$

Solution 1:

Notice that in the interval $[0, \pi/4]$ we have that $0 \leq \tan x \leq 1$. But then we have that \begin{align*} \tan^{n+1}(x) &\leq \tan^n(x) \\ \int_0^{\pi/ 4}\tan^{n+1}(x)dx &\leq \int_0^{\pi/ 4}\tan^{n}(x)dx \\ W_{n+1} &\leq W_n \end{align*} Now, I'll prove that $$ \lim_{n \to \infty} \frac{W_n}{\frac{1}{n}} = \frac{1}{2} $$ Using the equality you derived, we have that \begin{gather*} \frac{2W_{n+2}}{\frac{1}{n+1}} \leq \frac{W_n + W_{n+2}}{\frac{1}{n+1}} = 1 \leq \frac{2W_n}{\frac{1}{n+1}} \\ \frac{1}{2(n+1)} \leq W_n \qquad \frac{1}{2(n+1)} \geq W_{n+2} \Rightarrow W_n \leq \frac{1}{2(n-1)} \\ \frac{1}{2(n+1)} \leq W_n \leq \frac{1}{2(n-1)} \end{gather*} Applying squeeze limit, we get the limit exist and is equal to $\frac{1}{2}$. Finally, we apply ratio test \begin{align*} \lim_{n \to \infty} \left| \frac{\frac{W_{n+1}x^{n+1}}{(n+1)^\alpha}}{\frac{W_{n}x^{n}}{n^\alpha}} \right| = |x| \lim_{n \to \infty} \frac{\frac{W_{n+1}}{\frac{1}{n+1}}}{\frac{W_{n}}{\frac{1}{n}}} \frac{n^{\alpha+1}}{(n+1)^{\alpha+1}} = |x| \end{align*} Thus, $R = 1$