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Finding whether the series $$\;\; \sum^{\infty}_{k=1}\frac{k}{k^2-\sin^2(k)}$$ is converges or Diverges

What i try

We know that $\sin^2(x)\leq 1$ for all real number.

So $$\sum^{\infty}_{k=1}\frac{k}{k^2-\sin^2(k)}\geq \sum^{\infty}_{k=1}\frac{k}{k^2-1}$$

How do i solve it . Help me please.

jacky
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  • Start at $k=2$ to avoid division by $0$. What is the behavior of $\sum_{k=2}^{\infty} \frac{k}{k^2-1}$? – Integrand Jun 11 '20 at 15:50
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    Thanks Integrand . From $$\sum^{\infty}{k=2}\frac{k}{k^2-1}\geq\sum^{\infty}{k=2}\frac{k}{k^2}=\sum^{\infty}_{k=1}\frac{1}{k}$$ is diverges using $p$ series test. Is i am right – jacky Jun 11 '20 at 15:54
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    Note that $\frac{k}{k^2-\sin^2(k)}\ge \frac1{2k}$ for $k\ge 1$. – Mark Viola Jun 11 '20 at 16:33
  • The lower bound must be $\sum_{k=1}^{\infty}\frac{k}{k^2\color{red}{+}1}$. – metamorphy Jun 12 '20 at 04:42

1 Answers1

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You can write:

$\sum\limits_{k=2}^\infty \frac{k}{k^2-1}=\sum\limits_{k=2}^\infty \frac{1}{2(k-1)}+\frac{1}{2(k+1)}=\frac{1}{2}\sum\limits_{k=2}^\infty \frac{1}{k-1}+\frac{1}{k+1}$

Can you see where this is going?

CoffeeArabica
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