Determine if the following series converges or diverges and justify your decision.

The series diverges. We will show this by showing that the terms do not go to 0. To that end,

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),

Now, we can apply L’Hopital’s rule since this is the indeterminate form ,

So, putting this all back together we have,

Hence, the terms in the series do not tend to 0, so the series cannot converge.

\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

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