I am studying for my exam tomorrow and have come across some problems I cannot get. I have put them below with what I have tried/thinking process behind. Thank you.
1. Sum from 1 to infinity of ( 1 / (n + 2^n) )
I feel like I was supposed to use the comparison test for this one, but I don't know whether I should compare it to 1/n or 1/ 2^n, or even 1 / n + 2^n. Which one?
2. Sum from 1 to infinity of (n! / n^n)
I remember in class that whenever I see n!, I should be thinking to use the ratio test in order to have some n! cancel out. So I did this and I got this: (an/bn) = (an * bn reciprocal)
(n+1)! * n^n
______ ________
(n+1)^(n+1) n!
So would the next step be to split the (n+1)! up into (n) and (n!)? If so you get (and where I am stuck):
lim n^2n (?)
n--> infinity __________
(n+1)^(n+1)
I get stuck here and don't really know how to evaluate that with l'hopitals rule.
Thanks again
In a comment we described a simple Comparison approach to the second question. Here is an alternative using the Ratio Test.
As in the post, the ratio $\frac{a_{n+1}}{a_n}$ of consecutive terms is $$\frac{(n+1)!}{(n+1)^{n+1}}\cdot \frac{n^n}{n!}.$$
Cancel an $n+1$ from top and bottom of the first fraction. Things simplify to $$\frac{n^n}{(n+1)^n}$$.
This is the reciprocal of $\left(\frac{n+1}{n}\right)^n$, that is, of $\left(1+\frac{1}{n}\right)^n$, which has limit $e$ as $n\to\infty$.
It follows that $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{1}{e}\lt 1$, so we have convergence.
Remark: One can exploit the expression of $e$ as a limit to show that the terms in the third problem blow up as $n\to\infty$.