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

I’ve come across this sum, and I need to test the convergence, but I really have no idea how to get started. This is the sum: $$ \sum_{n=1}^{\infty} \frac{n^{\sqrt{6}+1}+n+5}{n^{\sqrt6+2}+2n+10} $$ Can you give me some tips? Thanks!.

share|cite|improve this question
up vote 6 down vote accepted

Hint: $\left(\dfrac{n^{\sqrt{6}+1}+n+5}{n^{\sqrt6+2}+2n+10}\right)_{n\in \mathbb N}\sim _\infty\left(\dfrac{n^{\sqrt{6}+1}}{n^{\sqrt6+2}}\right)_{n\in \mathbb N}=\,?$

share|cite|improve this answer
I get it, with comparision test my sum does not converge. Thank you. – user137209 Apr 27 '14 at 9:57
@user137209 No need to hurry in accepting my answer. Maybe you like another one better. – Git Gud Apr 27 '14 at 9:59

For $n\ge1$, $$ \begin{align} \frac{n^{\sqrt{6}+1}+n+5}{n^{\sqrt6+2}+2n+10} &=\frac1n\frac{1+n^{-\sqrt6}+5n^{-\sqrt6-1}}{1+2n^{-\sqrt6-1}+10n^{-\sqrt6-2}}\\ &\ge\frac1n\cdot\frac7{13} \end{align} $$

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.