1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Your argument is correct. $\endgroup$ – mfl Dec 2 '14 at 22:27
1
$\begingroup$

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, $$\sum_{n=1}^\infty \frac{(1+\frac{1}{n})^nn^2-7n}{n^3+3n^2+1} \geq \sum_{n=1}^\infty \frac{n^2-7n}{n^3+3n^2+1} \\ \geq \sum_{n=1}^\infty \frac{n^2-7n}{n^3+3n^3+n^3} \\ = \sum_{n=1}^\infty \frac{n^2-7n}{5n^3} \\ = \sum_{n=1}^\infty \frac{1}{5n}-\sum_{n=1}^\infty\frac{7}{5n^2} \\ = \frac{1}{5}\sum_{n=1}^\infty \frac{1}{n}-\frac{7\pi^2}{30}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can you show how $\sum_{n=1}^\infty \frac{7}{5n^2} = \sum_{n=1}^\infty \frac{7 \pi}{30} $? $\endgroup$ – bijonne Dec 3 '14 at 2:19
  • 1
    $\begingroup$ 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 $1$. $\endgroup$ – graydad Dec 3 '14 at 2:25

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.