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I have to check if this converges or diverges using the ratio test which is the $\lim_{n\to\infty} | a_{n+1} / a_{n} | $.

The problem is :

$$\sum_{n=1}^{\infty} \frac{\sin\frac1n}{\sqrt n}.$$

So far I tried:

$$\lim_{n→\infty} \left|\frac{\sin(1/(n+1))}{\sqrt{n+1}}\cdot\frac{\sqrt{n}}{\sin(1/n)}\right|.$$

After this step, I am stuck.

Please can someone help me?

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  • $\begingroup$ I think the ration test is inconclusive in this case. Why do you "hyave to use" that test? $\endgroup$ – DonAntonio Apr 17 '16 at 18:54
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The ratio test seems inconclusive to me. Indeed, looking at your ratio:

$$\frac{\sin\frac1{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{\sin\frac1n},$$

and noting that for $n\to\infty$, $\sin\frac1n=\frac1n+o(\frac1n)$, we see it is:

$$\frac{\frac{1}{n+1}+o\left(\frac{1}{n+1}\right)}{\sqrt{n+1}}\frac{\sqrt{n}}{\frac1n+o\left(\frac1n\right)}\to1,$$

but if the ratio tends to one the ratio test is inconclusive.

However, as noted in another answer (albeit with less details as to why the comparability holds), this series is comparable to $\sum\frac{1}{n^{^3/_2}}$, which is what is obtained by using the fact $\sin\frac1n\sim\frac1n$ as $n\to\infty$ directly in the sum. So it converges, since it has the same behaviour as that sum, which is $\frac{1}{n^p}$ for $p>1$, a family of series known to converge.

By the way, the root test is also inconclusive according to Wolfram. Comparison test is basically what I did. Ratio test isn't even considered by Wolfram… interesting :).

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Hint : $\sin(\frac{1}{n}) \approx \frac{1}{n}$ when $n$ tends to $\infty$. Can you start from there ?

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  • $\begingroup$ yes, i tried with sin(1/n+1) = 1 and sin (1/n) = 1 but then the answer should be 3/2 $\endgroup$ – bryan Apr 17 '16 at 18:43
  • $\begingroup$ What are the egalities in your comment ? It seems wrong. $\endgroup$ – Jennifer Apr 17 '16 at 18:45
  • $\begingroup$ @bryan I honestly have trouble understanding what you tried: your comment makes basically no sense, because why should you try to find $\sin(1/n)=1$ or $\sin(1/(n+1))=1$, and how could you possibly have them equal, and where does $3/2$ come from? $\endgroup$ – MickG Apr 17 '16 at 18:56
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The series is comparable to $\sum\frac{1}{n^{^3/_2}}$ which converges.

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$$\frac{a_{n+1}}{a_n}=\frac{\sin\frac1{n+1}}{\sqrt{n+1}}\cdot\frac{\sqrt n}{\sin\frac1n}=\frac{\sin\frac1{n+1}}{\frac1{n+1}}\cdot\frac{n}{n+1}\cdot\frac{\frac1n}{\sin\frac1n}\cdot\frac{\sqrt n}{\sqrt{n+1}}\xrightarrow[n\to\infty]{}1\cdot1\cdot1\cdot1=1$$

So the ratio test is inconclusive.

But we have that

$$\frac{\sin\frac1n}n\le\frac1{n\sqrt n}\iff\sqrt n\sin\frac1n\le1\;\;\color{red}{(**)}$$

and we have that

$$\sqrt n\sin\frac1n=\frac1{\sqrt n}\cdot\frac{\sin\frac1n}{\frac1n}\xrightarrow[n\to\infty]{}0\cdot1=0$$

so the inequality $\;\color{red}{(**)}\;$ above is true for $\;n\;$ big enough, and thus the comparison test gives us the convergence of our series.

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  • $\begingroup$ that was helpfull, thank you $\endgroup$ – bryan Apr 17 '16 at 19:13
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why don't you try that putting (-1)^n instead of sin(1/n) and then your series will be $\sum_{n=0}^\infty \frac{(-1)^n}{\sqrt{n}}$ and when you apply ratio test you will be fail but we know that this series divergent by p-test,however immadietely see that it is conditionally convergent by alternating series test.

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You can in fact show that

$$\sum_{n=0}^{\infty} \frac{1}{n^p}\sin\left(\frac{1}{n^q}\right)$$

converges iff $p+q>1$.

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