3
$\begingroup$

Does the following series converge or diverge $$\sum_{n=1}^{\infty}\frac{1}{n(\ln{n})^2+n}$$ I know that $\sum_{n=1}^{\infty}\frac{1}{(\ln{n})^2}$ diverges.

$\sum_{n=1}^{\infty}\frac{1}{n(\ln{n})^2}$ dominates the series in question, and converges. What comparison can I use to understand this?

$\endgroup$
  • $\begingroup$ Useful keyword: it's a Bertrand series. $\endgroup$ – Clement C. Jun 20 '16 at 15:34
  • $\begingroup$ Show that the convergence of your series is implied by the convergence of $\sum_{n=2}^{\infty} \frac 1{n(\ln n)^2}$. Then use integral test. $\endgroup$ – Sungjin Kim Jun 20 '16 at 16:05
6
$\begingroup$

Note that

$$\frac{1}{n\log^2(n)+n}\le \frac{1}{n\log^2(n)}$$

and

$$\int_2^\infty \frac{1}{x\log^2(x)}\,dx=\frac{1}{\log(2)}$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The Cauchy condensation test gives that your series is convergent iff $$ \sum_{m\geq 0}\frac{1}{1+m^2}=\frac{1+\pi\coth \pi}{2} $$ is convergent.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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