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I would like to prove if the next two series are convergent. First: $$ \sum_{n=1}^{\infty}\log\left(\frac{n+1}{n}\right)\arcsin \left(\frac{1}{\sqrt{n}}\right) $$ I think that this series is convergent, so $$\arcsin\left(\frac{1}{\sqrt{n}}\right)$$ is similar to $$\frac{1}{\sqrt {n}}$$. And $$\log\left(\frac{n+1}{n}\right)=\log\left(1+\frac{1}{n}\right)\sim \frac {1}{n} $$ if n goes to infinity. So I have the series $$\sum_{n=1}^{\infty}\frac{1}{n}\frac{1}{\sqrt {n}}$$, this series converge. Is this argument valid to prove the convergence? Second: $$\sum_{n=1}^{\infty}1-\sec\left(\frac{1}{n}\right)$$. Could you help me please? Give me some clue please!!!!

Thank you

Why Would You
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2 Answers2

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I can give you a hint for the first one. I think that your intuition is correct. To make it rigorous you can either write the logarithm and arcsin as Taylor series until the first degree + the error (given by a big O of something, or an explicit integral) and look at what it happens when you distribute the sum, or you can try to approximate the series by an upper integral, an integrate by parts (personally I think that the first approach is way simpler).

Duca_Conte
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Answer for the second series: this series converges absolutely if $\sum \frac {1-\cos(\frac 1 n)} {|\cos(\frac 1 n)|}$ converges. Since the denominator tends to $1$ it is enough to prove convergence of $\sum {(1-\cos(\frac 1 n))}$. This series converges because $1-\cos \theta \leq \frac {\theta^{2}} {2}$ and $\sum \frac 1 {n^{2}} <\infty$. [ Proof of $1-\cos \theta \leq \frac {\theta^{2}} {2}$: $1-\cos \theta =2\sin^{2} \frac {\theta} 2 \leq \frac {\theta^{2}} {2}$ since $|\sin t| \leq t$ for all $t \geq 0$].

Kavi Rama Murthy
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