Determine if the following series converges or diverges and justify your decision.
![\[ \sum_{n=1}^{\infty} \frac{n^{n+\frac{1}{n}}}{\left( n + \frac{1}{n} \right)^n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f77e273cdfe87834da9abd4517f3446d_l3.png "Rendered by QuickLaTeX.com")
The series diverges. We will show this by showing that the terms 
do not go to 0. To that end,
![\begin{align*} \lim_{n \to \infty} a_n &= \lim_{n \to \infty} \frac{n^{n + \frac{1}{n}}}{\left( n + \frac{1}{n} \right)^n} \\[9pt] &= \lim_{n \to \infty} \frac{e^{\left( n + \frac{1}{n} \right) \log n}}{e^{n \log \left( n + \frac{1}{n} \right)}} \\[9pt] &= \exp \left( \lim_{n \to \infty} \left( \left( n + \frac{1}{n} \right) \log n - n \log \left( n + \frac{1}{n} \right) \right) \right) \\[9pt] &= \exp \left( \lim_{n \to \infty} \left( n (\log n - \log(n+1/n)) + \frac{\log n}{n} \right) \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e2ba5445eeac779e34a7a47f2fc5fe03_l3.png "Rendered by QuickLaTeX.com")
So, we’d like to split this limit of a sum into a sum of limits, which we can do if the each of the limits exist on their own. We know 
, so we need to worry about the first term. We compute its value (and the fact that it’s finite will justify splitting the limit into two separate limits),
![\begin{align*} \lim_{n \to \infty} \left( n (\log n - \log (n + 1/n)) \right) &= \lim_{n \to \infty} \left( n \log \left( \frac{n}{n+\frac{1}{n}} \right) \right) \\[9pt] &= \lim_{n \to \infty} \frac{\log \left( \frac{1}{1+\frac{1}{n^2}} \right)}{\frac{1}{n}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-83e737e9dfe44e0b942ec45ac0490501_l3.png "Rendered by QuickLaTeX.com")
Now, we can apply L’Hopital’s rule since this is the indeterminate form 
,
![\begin{align*} &= \lim_{n \to \infty} \left( -\frac{1}{n^2} \right)^{-1} \cdot \frac{2}{n^3+n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{-2n}{n^2+1} \\[9pt] &= 0. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-511a9fee8f99ba8ba02874bfd1c4711b_l3.png "Rendered by QuickLaTeX.com")
So, putting this all back together we have,
![\begin{align*} \lim_{n \to \infty} a_n &= \exp \left( \lim_{n \to \infty} \left( n (\log n - \log(n+1/n)) + \frac{\log n}{n} \right) \right) \\[9pt] &= \exp \left(\lim_{n \to \infty} \left(n (\log n - \log (n+1/n))\right) + \lim_{n \to \infty} \frac{\log n}{n} \right) \\[9pt] &= e^{0 + 0} \\[9pt] &= 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-60ceb614fccc78dec7a2b6ca7a99716d_l3.png "Rendered by QuickLaTeX.com")
Hence, the terms in the series do not tend to 0, so the series cannot converge.
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A simpler way would be to find a different lower bound which would diverge. Here the bound would be:
. Then the limit of its term sequence shows they tend to 1. The limit can be computed with the Binomial theorem decomposition.
\lim_{n\to\infty}\frac {n^{n＋1/n}}{(n＋1/n)^n}＝\lim_{n\to\infty}\left(\frac {n}{n＋1/n}\right)^nn^{1/n}≥\lim_{n\to\infty}\left(\frac {n}{n＋1}\right)^nn^{1/n}＝\frac {1}{e}>0
I like how you used L’Hopital’s rule here but a simpler method may suffice: n^(n+1/n)/(n+1/n)^n < n^(n+1/n)/n^n = n^(1/n), which does not go to zero.
That doesn’t imply anything, since being less than a term which goes to one doesn’t mean that the terms don’t go to zero