# Did I correctly verify the convergence of this series?

I want to find if the following series is convergent.

$$\sum_{n=1}^\infty \frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1}$$

I use the asymptotic criterion for series convergence.

$$a_n=\frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1}$$

I take such $b_n$ that $a_n$ and $b_n$ are asymptotically similar and that the convergence of $\sum_{n=1}^\infty b_n$ is known.

$$b_n=\frac{1}{n}$$

$$\lim_{n\to \infty}\frac{a_n}{b_n}=\lim_{n\to \infty}\frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} \frac {n}{1}=\lim_{n\to \infty}\frac{(1+\frac{1}{n})^nn^3-7n^2}{n^3+3n^2+1}$$

The limit is $e$ which proves that $a_n \sim b_n$.

Then since $\sum_{n=1}^\infty \frac{1}{n}$ is divergent, so is the original series.

I'd be thankful if someone could review this and tell me if this solution is correct.

• Your argument is correct. – mfl Dec 2 '14 at 22:27

Your conclusion is correct. You could get there using the comparison test too if you are interested. Namely, that $$1 < \left(1+\frac{1}{n} \right)^n \\ \frac{1}{n^3+3n^3+n^3}\leq \frac{1}{n^3+3n^2+1}$$ for all $n \geq 1$. Hence, $$\sum_{n=1}^\infty \frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} \geq \sum_{n=1}^\infty \frac{n^2-7n}{n^3+3n^2+1} \\ \geq \sum_{n=1}^\infty \frac{n^2-7n}{n^3+3n^3+n^3} \\ = \sum_{n=1}^\infty \frac{n^2-7n}{5n^3} \\ = \sum_{n=1}^\infty \frac{1}{5n}-\sum_{n=1}^\infty\frac{7}{5n^2} \\ = \frac{1}{5}\sum_{n=1}^\infty \frac{1}{n}-\frac{7\pi^2}{30}$$
• Can you show how $\sum_{n=1}^\infty \frac{7}{5n^2} = \sum_{n=1}^\infty \frac{7 \pi}{30}$? – bijonne Dec 3 '14 at 2:19
• Sadly I can't do much better than to say it is because $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$ hence if you multiply by $\frac{7}{5}$ you will get the result. But the fact that $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$ comes from the Riemann-Zeta function. It's not something you'd likely encounter until you have covered quite a bit of complex analysis. The Riemann-Zeta function is a useful tool to determine the value of sums of the form $$\sum_{n=1}^\infty \frac{1}{n^s}$$ where $s$ is any complex number not equal to $1$. – graydad Dec 3 '14 at 2:25