Determine whether the following is converging or diverging

$$\sum_{i=1}^∞ \frac{\sin(1/i)}{\sqrt{i}}$$

I tried the following, but not sure whether it's correct:

$$\frac{\sin(1/i)}{\sqrt i}$$ <= $$\frac {1}{n\sqrt n}$$

Hence by the nth term test, $$\frac {1}{n\sqrt n}$$ is convergent, hence the series converges

But I am not sure how to prove $$\frac{\sin(1/i)}{\sqrt i}$$ is less than or = to $$\frac {1}{n\sqrt n}$$

  • 2
    $\begingroup$ You do not need to evaluate it to prove it is a convergent series. The (asymptotic) comparison test and the p-test are enough. $\endgroup$ Mar 29, 2018 at 11:47
  • $\begingroup$ I was in the process of making the same comment as Jack when I saw his. So let me just elaborate: evaluate means find the value of. $\endgroup$ Mar 29, 2018 at 11:49

2 Answers 2


Hint. Every term of your sum is non-negative. Use the inequality $\sin x\le x$ for $x\ge 0$.

  • $\begingroup$ Can i check that the above series is divergent? $\endgroup$
    – Isabella.T
    Mar 29, 2018 at 12:52
  • $\begingroup$ @Isabella.T No, your series is converges. (Why?) $\endgroup$
    – Hanul Jeon
    Mar 29, 2018 at 23:07

Simply note that

$$\frac{\sin(1/i)}{\sqrt{i}}\sim \frac{1}{i\sqrt{i}}$$

then the given series converges by limit comparison test with $\sum \frac{1}{i\sqrt{i}}$.

  • $\begingroup$ may i understand how do you get that sin(1/i)/sqrt(i) <= 1/isqrt(i)? $\endgroup$
    – Isabella.T
    Mar 30, 2018 at 1:25
  • $\begingroup$ @Isabella.T Since for $$\frac1i \to 0 \implies \sin\frac1i \sim \frac1i$$ $\endgroup$
    – user
    Mar 30, 2018 at 3:56

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