Solved: "Convergence of the series∑n=1∞(1nα)(∫0π4tanntdt)xn"

Junaid Ayala

Answered question

2022-02-25

Convergence of the series

\sum_{n = 1}^{\infty} (\frac{1}{n^{α}}) (\int_{0}^{\frac{π}{4}} \tan^{n} t d t) x^{n}

Brody Buckley

Beginner2022-02-26Added 5 answers

Notice that in the interval

[0, \frac{π}{4}]

we have that

0 \leq \tan x \leq 1

. But then we have that

\tan^{n + 1} (x) \leq \tan^{n} (x)

\int_{0}^{\frac{π}{4}} \tan^{n + 1} (x) d x \leq \int_{0}^{\frac{π}{4}} \tan^{n} (x) d x

W_{n + 1} \leq W_{n}

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

\frac{2 W_{n + 2}}{\frac{1}{n + 1}} \leq \frac{W_{n} + W_{n + 2}}{\frac{1}{n + 1}} = 1 \leq \frac{2 W_{n}}{\frac{1}{n + 1}}

\frac{1}{2 (n + 1)} \leq W_{n} \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)}

Applying squeeze limit, we get the limit exist and is equal to

\frac{1}{2}

. Finally, we apply ratio test

lim_{n \to \infty} | \frac{\frac{W_{n + 1} x^{n + 1}}{{(n + 1)}^{a}}}{\frac{W_{n} x^{n}}{n^{a}}} | = | x | lim_{n \to \infty} \frac{\frac{W_{n + 1}}{\frac{1}{n + 1}}}{\frac{W_{n}}{\frac{1}{n}}} \frac{n^{a + 1}}{{(n + 1)}^{a + 1}} = | x |

Thus,

R = 1

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