Originally Posted by
angel.white Comparison test:
let $\displaystyle a_n = \frac{arctan(n)}{n^{1.2}}$, and let $\displaystyle b_n = \frac{\pi}{n^{1.2}}$
Then $\displaystyle a_n \leq b_n$ (because $\displaystyle \lim_{n\to\infty} arctan(n) = \frac {\pi}2$)
And $\displaystyle b_n$ is a p-series with p=1.2, which is greater than 1, therefore $\displaystyle \sum b_n$ converges, and because $\displaystyle \sum a_n$ is monotonic and increasing, and bounded above by $\displaystyle \sum b_n$ we know that $\displaystyle \sum a_n$ converges by the comparison test.
Ratio Test:
$\displaystyle a_n =\frac{n^n}{(2n)!}$, and $\displaystyle a_{n+1} = \frac{(n+1)^{n+1}}{(2(n+1))!} = \frac{(n+1)^{n+1}}{(2n+1)(2n+2)(2n)!}$
$\displaystyle \lim_{n\to\infty} \left| \frac {a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left| \frac{(n+1)^{n+1}}{(2n+1)(2n+2)(2n)!} * \frac{(2n)!}{n^n}\right|$
$\displaystyle =\lim_{n\to\infty} \left| \frac{(n+1)^{n+1}}{(2n+1)(2n+2)n^n}\right|$
$\displaystyle =\lim_{n\to\infty} \left| \frac{(n+1)(n+1)^n}{(2n+1)(2n+2)n^n}\right|$
Here multiply numerator and denominator by 1 over n^n to get
$\displaystyle =\lim_{n\to\infty} \left| \frac{(n+1)(1+1/n)^n}{(2n+1)(2n+2)}\right|$
Now multiply the numerator and denominator by 1/n to get
$\displaystyle =\lim_{n\to\infty} \left| \frac{(1+1/n)(1+1/n)^n}{(2+1/n)(2n+2)}\right|$
It should be relatively clear that this converges to zero, (because the numerator converges to e, and the denominator diverges) but if you need me to explain that more, I can.
So because $\displaystyle \lim_{n\to\infty} \left| \frac {a_{n+1}}{a_n}\right| = 0 < 1$ the series $\displaystyle \sum_n^\infty \frac{n^n}{(2n)!}$ is absolutely convergent and therefore convergent.