"Proof that ∑(−1)nan converge with an=(n+2−n+1)sin⁡1n" was answered by our community

Dachpunkt4cr

Answered question

2022-02-22

Proof that

\sum {(- 1)}^{n} a_{n}

converge with

a_{n} = (\sqrt{n + 2} - \sqrt{n + 1}) \sin \frac{1}{n}

Malaika Ridley

Beginner2022-02-23Added 6 answers

Using Taylor expansion for large values of n, you could show that

\frac{1}{2 n^{\frac{3}{2}}} - \frac{3}{8 n^{\frac{5}{2}}} < [\sqrt{n + 2} - \sqrt{n + 1}] \sin (\frac{1}{n}) < \frac{1}{2 n^{\frac{3}{2}}}

and then

\frac{1}{2} ζ (\frac{3}{2}) - \frac{3}{8} ζ (\frac{5}{2}) < \sum_{n = 1}^{\infty} [\sqrt{n + 2} - \sqrt{n + 1}] \sin (\frac{1}{n}) < \frac{1}{2} ζ (\frac{3}{2})

Numerically,

0.80313 < \sum_{n = 1}^{\infty} [\sqrt{n + 2} - \sqrt{n + 1}] \sin (\frac{1}{n}) < 1.30619

while the summation is 0.97329

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