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Could somebody please check my solution?

I want to check, whether $\sum\limits_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{n^2}$ converges or diverges.

Using the Comparison test:

Let $a_n = \frac{(1+\frac{1}{n})^n}{n^2},~ b_n=\frac{1}{n^2}$

Since $\sum\limits_{n=1}^{\infty}\frac{1}{n^2}$ converges and $\lim\limits_{n \rightarrow \infty} \frac{a_n}{b_n}= \lim\limits_{n \rightarrow \infty} \frac{(1+\frac{1}{n})^n}{n^2} \frac{n^2}{1} = \lim\limits_{n \rightarrow \infty} (1+\frac{1}{n})^n = e$.

Since $0<e<\infty$, $\sum\limits_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{n^2}$ converges.

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    $\begingroup$ It does work, but wasn't is simpler to notice that $$ 0\leq \frac{(1+\frac1n)^n}{n^2}\leq \frac{e}{n^2} \text? $$ $\endgroup$
    – AndreasT
    Commented Dec 25, 2013 at 22:58
  • $\begingroup$ It is correct. Nice work. $\endgroup$ Commented Dec 25, 2013 at 22:59
  • $\begingroup$ Your solution is correct. $\endgroup$ Commented Dec 25, 2013 at 23:00
  • $\begingroup$ Good and simple answer. $\endgroup$
    – Semsem
    Commented Dec 26, 2013 at 0:21
  • $\begingroup$ I do not see why someone would want to add the word limit in front of the comparison test? I think one should respect the OP's language/notation $\endgroup$
    – Lost1
    Commented Dec 26, 2013 at 1:43

2 Answers 2

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Your solution is correct. However, the simpler trick implied by the problem is to notice the numerator. Recall that $$ \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n=e $$ Since all the terms are positive, we have $$ 0\leq \sum_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^n}{n^2} \leq \sum_{n=1}^\infty \frac{e}{n^2}= e \sum_{n=1}^\infty \frac{1}{n^2} $$ By the $p$-test, the series on the right converges. Therefore, your original series converges.

EDIT As suggested by user21820, it might be unclear what I was saying. The limit just produces me a number, $e$ to use to bound the original series. The solution 'works' because $\left(1+\frac{1}{n}\right)^n<e$ for any positive integer $n$. This is usually discussed when one learns the definition of $e$. So then each term on the right sum is larger than that of corresponding ones in the original. The use of $e$ was arbitrary (but suggested by the numerator). We could have used 'any' number, $\pi, e^2,4,10,\sqrt{17}$, so long as $\left(1+\frac{1}{n}\right)^n$ was smaller than our choice of number, say $x$, for all positive integer $n$ (or at least all but finitely many of them). Then we would have written $$ 0\leq \sum_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^n}{n^2} \leq \sum_{n=1}^\infty \frac{x}{n^2}= x \sum_{n=1}^\infty \frac{1}{n^2} $$ for whatever larger number we chose.

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  • $\begingroup$ Hmm did you forget to mention that $(1+\frac{1}{n})^n<e$ for any positive integer $n$? If you just use the limit being $e$, you need to ignore a finite number of terms and add in some little epsilon somewhere. $\endgroup$
    – user21820
    Commented Dec 26, 2013 at 1:42
  • $\begingroup$ @user21820 You don't need to ignore any number of terms, we aren't looking at partial sums in my answer. Instead, we are looking at the series as a whole. Moreover, even when looking at a finite number of terms, as you point out, $\left(1+1/n\right)^n<e$ for any positive integer $n$. Because of this, $e$ in the numerator is larger than $(1+1/n)^n$ for any positive integer $n$ in the numerator. Hence, even when looking at partial sums, the series on the right in my answer is always larger than the partial sums of the original series. No epsilon argument or ignoring terms needed. $\endgroup$ Commented Dec 26, 2013 at 2:27
  • $\begingroup$ No in your answer you stated the limit of $(1+\frac{1}{n})^n$ to be $e$, but that does not give what you want without using what I described. Instead you need the actual inequality as I said. $\endgroup$
    – user21820
    Commented Dec 26, 2013 at 2:28
  • $\begingroup$ @user21820 I believe you are confused. I use the fact that $\lim_{n\to \infty}(1+1/n)^n$ is $e$. Since $e>(1+1/n)^n$ for all positive integer $n$, $e/n^2>(1+1/n)^n/n^2$ for all positive integer $n$. Then whether you are looking at partial sums or the series as a whole, you have the required inequality. I'm not trying to show that in the limit the original sequence becomes 'like' $\sum e/n^2$. Rather, I'm just finding something bigger to bound the original series by. I could have equally used $\sum e^2/n^2$ or $\sum 4/n^2$ or $\sum \pi/n^2$. $\endgroup$ Commented Dec 26, 2013 at 2:32
  • $\begingroup$ That is why I said you should have mentioned the inequality because it was nowhere in your answer, and readers might get the incorrect impression that the limit is enough to justify your later bounding. The second part of my first comment was to say that you could indeed do with just the limit, except that you need more work. I mentioned that because in some cases we don't have a strict inequality like we do in this case, but the alternative technique remains valid. $\endgroup$
    – user21820
    Commented Dec 26, 2013 at 2:37
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Using the limit comparison test, $$ {1\over n^2}\left(1 + {1\over n}\right)^n \sim {e\over n^2},$$ your series converges.

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