Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$ \sum_{i=1}^\infty {2n+3\over 5n+1} $$

$$ \sum_{i=1}^\infty \left({2n+3\over 5n+1}\right)^n $$

$$ \sum_{i=1}^\infty {\sin^2n\over n \sqrt{n}} $$

for 1st one is it okay to apply divergence test?

what is the test for others also?

thanks in advance

share|cite|improve this question
For the first, yes, the divergence test is the test to apply. For the second series, apply the root test, and for the third, compare it with $1/n\sqrt{n}$ and check convergence using $p$-series. – Clayton Jan 4 '13 at 2:05

First $a_n\to ?$

Second $(a_n)^{1/n}\to ?$

Third does $\int a_n$ converge?

share|cite|improve this answer
actually I didn't understand your answer. – Mustafa Jan 4 '13 at 2:12
@Okan In the first case I ask you to find where the $n$-th term of the series tends. Knowing this answer you will answer the question of the series convergence. Others are similar. – Artem Jan 4 '13 at 2:15
For the third one, you can't use integral test since it's not monotonic. Anyway, you can compare it to $\sum \frac{1}{n^{3/2}}$. – Amihai Zivan Jan 4 '13 at 7:13

Notice that if the terms don't go to zero, the series can't converge. For the third one, there's a simple comparision test to a series I expect you know about. For the second, compare to the sum of $(2/5)^n$.

share|cite|improve this answer

About the thrid one, you can also use this fact that $$\lim_{n\to\infty}n^{\frac{-1}{2}}\frac{\sin^2(n)}{n\sqrt{n}}=1\neq$$ so it diverges. For others @Artem's answer is practical.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.