does $\sum_{n=1}^\infty n!/(n^n+5^n)$ converge? I am asked to determine if this converges, using the ratio test I got $n^n/(5(n+1)^n)<1$ so its decreasing.
This is where I am stuck, I think 0 is a lower bound so that by the monotone convergence theorem it is convergent but I am unsure how to prove this.
The question also hints to use both the comparison and ratio tests but I can't get the comparison to work.
 A: Compare to $a_n = \frac{n!}{n^n}$. Now use the ratio test:
$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)!n^n}{n! (n+1)^{(n+1)}} = \left(\frac{n}{n+1}\right)^n $$
which limits to $1/e < 1$.
How did I know to do this? Three tips:

*

*Always start by eyeballing the term. Which factors dominate the numerator and denominator? Which are insignificant in the limit? In this case, $n^n$ grows much much faster than $5^n$, so our intuition should be: we can "throw away" $5^n$ and have a limit with the same behavior.

*Use the comparison test to simplify, simplify, simplify. Based on the above, we'd like to throw away $5^n$ for a simpler series... and that's what the comparison test lets us do. (Exercise: What would you do if we were subtracting $5^n$ from the denominator instead of adding it?)

*Both $n!$ and $n^n$ are defined multiplicatively, so that means the ratio test may be a good way to attack them.

After that, it was a matter of some algebra to check these guesses, and then recognizing the (disguised) limit definition of $e$.
A: Letting $a_n$ be the terms in the series; that is $a_n=\frac{n!}{n^n+5^n}$, from the ratio test, you get
$$\frac{a_{n}}{a_{n-1}}=\frac{\frac{n!}{n^{n}+5^{n}}}{\frac{(n-1)!}{(n-1)^{n-1}+5^{n-1}}}=\frac{n\left((n-1)^{n-1}+5^{n-1}\right)}{n^n+5^n}$$
This is less than $$\frac{n\left((n-1)^{n-1}+5^n\right)}{n^n}=\left(\frac{n-1}{n}\right)^{n-1}+5\left(\frac{5}{n}\right)^{n-1} \to \left(1-\frac{1}{n}\right)^{n-1}+0\to\frac{1}{e}$$
Since this limit is less than $1$, the series converges by the ratio test.
