How prove this series is converges? $\sum_{n=1}^{\infty}\dfrac{1}{n\ln{(n^3+n)}}$

Question

this series $$\sum_{n=1}^{\infty}\dfrac{1}{n\ln{(n^3+n)}}$$ is converge ？

I conside $$\dfrac{x}{1+x}<\ln{(1+x)}<x$$ then $$\ln{(n^3+n)}\ge\dfrac{n^3+n-1}{n^3+n}$$ then $$\dfrac{1}{n\ln{(n^3+n)}}<\dfrac{n^3+n}{n(n^3+n-1)}$$ then I can't,Thank you

Answer 1

Hint: Use $n^3<n^3+n<n^4$ for $n>1$. Then show this series converges if and only if $\sum_{n=2}^\infty \frac{1}{n\ln n}$ converges.

Answer 2

The series is, in fact, divergent.

Note that $n^3 + n \le n^4$ for large $n$, so that

$$\ln (n^3 + n) \le \ln(n^4)$$

Therefore,

$$\frac{1}{\ln(n^3 + n)} \ge \frac{1}{\ln n^4} = \frac 1 {4 \ln n}$$

Now consider the series

$$\sum_{n = 2}^{\infty} \frac 1 {4n \ln n}$$

By the integral test, and using the fact that

$$\int \frac{1}{x \ln x} dx = \ln \ln x + c$$

we see that this series is divergent. The original series is then divergent by comparison.

Answer 3

I would try to make the terms larger but removing the $ n $ from inside the log. Since $\ln (n^3 + n) > \ln (n^3) $, the reciprocals reverse the inequality. Then try a comparison integral.