# Does the series $\sum_{n=1}^\infty \frac{n+1}{n^3+10n}$ converge?

Using the ratio test, we evaluate: $$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty}\left| \frac{(n+1) + 1}{(n+1)^3 + 10(n+1)} \cdot \frac{n^3+10n}{n+1} \right| = \lim_{n \to \infty} \left| \frac{n^4 + 2n^3 + 10n^2 + 20n}{(n+1)^4 + 10(n+1)^2} \right|$$ $$< \lim_{n \to \infty} \left| \frac{n^4 + 2n^3 + 10n^2 + 20n}{n^4 + 10n^2} \right| = 1$$

Hence $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$ and the series converges.

Is this an appropriate solution?

• No. The critical step (the $<$ step) in your proof is not obvious; in fact, it is false and both limits are $1$. You need to use the comparison test against $\frac{1}{n^2}$. – rogerl Oct 13 '14 at 21:28

We observe first that $$\frac{n+1}{n^3+10n}\le \frac{n+1}{n^3+n}=\frac{1}{n^2},$$ and as the series $$\sum_{n=1}^\infty\frac{1}{n^2},$$ converges, so does the series $\,\,\,\displaystyle\sum_{n=1}^\infty\frac{n+1}{n^3+10n}$.
You can't conclude using the ratio test since the limit is $1$. Notice that
$$a_n\sim_\infty \frac1{n^2}$$ and the Riemann series $\sum\frac1{n^2}$ is convergent so the given series is convergent by comparison.
The series is bounded above by $$\sum \frac{n + 1}{n^3} = \sum \frac{1}{n^2} + \sum \frac{1}{n^3},$$ so by the comparison test the series converges because both sums on the right-hand side are convergent $p$-series.
Hint. You may use the comparison test, writing $$\frac{n+1}{n^3+10n}=\frac{1+\frac1n}{n^2+10}\sim \frac{1}{n^2}$$ as $n$ tends to $+\infty$ and conclude easily.
$$\sum_{n=1}^\infty \frac{n+1}{n^3+10n} \leq \sum_{n=1}^\infty \frac{n+n}{n^3} = 2 \sum_{n=1}^\infty \frac{1}{n^2} < \infty.$$