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$$\sum\limits_{n=1}^\infty\frac{\sin(\sqrt n)}{n^{\frac{3}{2}}}$$ Let ${a_n} = \frac{{\sin (\sqrt n )}}{{{n^{\frac{3}{2}}}}}$, then, using the criterion of quotient I must prove that $\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|<1$, indeed: $$\begin{array}{l} \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\frac{{\sin (\sqrt {n + 1} )}}{{{{(n + 1)}^{\frac{3}{2}}}}}}}{{\frac{{\sin (\sqrt n )}}{{{n^{\frac{3}{2}}}}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\sin (\sqrt {n + 1} ) \cdot {n^{\frac{3}{2}}}}}{{{{(n + 1)}^{\frac{3}{2}}} \cdot \sin (\sqrt n )}}} \right|\\ = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\sin (\sqrt {n + 1} ) \cdot {n^{\frac{3}{2}}}}}{{{{(n + 1)}^{\frac{3}{2}}} \cdot \sin (\sqrt n )}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\sin (\sqrt {n + 1} )}}{{\sin (\sqrt n )}}} \right|{\left| {\frac{n}{{n + 1}}} \right|^{\frac{3}{2}}}\\ = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\sin (\sqrt {n + 1} )}}{{\sin (\sqrt n )}}} \right| \cdot \mathop {\lim }\limits_{n \to \infty } {\left| {\frac{n}{{n + 1}}} \right|^{\frac{3}{2}}} = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\sin (\sqrt {n + 1} )}}{{\sin (\sqrt n )}}} \right| \cdot (1)\\ = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\sin (\sqrt {n + 1} )}}{{\sin (\sqrt n )}}} \right| \end{array}$$ then I stay stagnant.

How I can know if this limit is $ \geq1 $ or $ < 1?$

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    $\begingroup$ You don't need to work so hard. Just use the fact that $|\sin(\sqrt{n})| \leq 1$ $\endgroup$
    – Prahlad Vaidyanathan
    Commented Dec 3, 2013 at 8:22
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    $\begingroup$ Do you really want to start at $n=0$? $\endgroup$
    – mrf
    Commented Dec 3, 2013 at 8:25
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    $\begingroup$ That last limit that gives you trouble probably doesn't exist. So don't use the criterion of quotient, just use the upper bound from Prahlad's comment. $\endgroup$
    – Dan Shved
    Commented Dec 3, 2013 at 8:25
  • $\begingroup$ If that limit is $=1$ then you don't know anything about convergence. Only if $\neq 1$ you know something. $\endgroup$
    – Cortizol
    Commented Dec 3, 2013 at 8:29
  • $\begingroup$ @mrf Sorry, the series star at $n=1$, I was wrong typing $\endgroup$
    – mathsalomon
    Commented Dec 3, 2013 at 8:33

1 Answer 1

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Let us take absolute series of given series, ie.$$\sum_{n=1}^\infty |\frac {\sin \sqrt n}{n^{3/2}}|$$.since we know $|\frac {\sin \sqrt n}{n^{3/2}}|$$\leq$ $|\frac {1}{n^{3/2}}|$.But $\sum_{n=1}^{\infty} |\frac {1}{n^{3/2}}|$ is converging.Hence by comparison theorem, $\sum_{n=1}^\infty |\frac {\sin \sqrt n}{n^{3/2}}|$ converges.Since absolute series conveges,so the given series converges... $$ $$ Hence proved

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  • $\begingroup$ Perfect! Thanks Jibin Joy K. Is clear, comparison theorem is very useful $\endgroup$
    – mathsalomon
    Commented Dec 3, 2013 at 8:39
  • $\begingroup$ Sure. I forgot to mention that I wanted to know the absolute convergence, not only the convergence. $\endgroup$
    – mathsalomon
    Commented Dec 3, 2013 at 9:40

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