Convergence/divergence series

Question

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 $$\frac {1}{n\sqrt n}$$

You do not need to evaluate it to prove it is a convergent series. The (asymptotic) comparison test and the p-test are enough. — Jack D'Aurizio, Mar 29 '18 at 11:47
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. — Matthew Leingang, Mar 29 '18 at 11:49

score 3 · Answer 1 · answered Mar 29 '18 at 11:44

3

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

answered Mar 29 '18 at 11:44

Hanul Jeon

Can i check that the above series is divergent? – Isabella.T Mar 29 '18 at 12:52
@Isabella.T No, your series is converges. (Why?) – Hanul Jeon Mar 29 '18 at 23:07

score 1 · Accepted Answer · answered Mar 29 '18 at 12:38

1

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 $\sum \frac{1}{i\sqrt{i}}$.

answered Mar 29 '18 at 12:38

user

may i understand how do you get that sin(1/i)/sqrt(i) <= 1/isqrt(i)? – Isabella.T Mar 30 '18 at 01:25
@Isabella.T Since for $$\frac1i \to 0 \implies \sin\frac1i \sim \frac1i$$ – user Mar 30 '18 at 03:56

2 Answers2