sum 1-->infinity of (arctan(1/n)^n
Here's what I've done:
lim 1->infinity of: nth root (above expression)
= lim 1-> infinity of arctan(1/n)
Now what?
At first the problem seems an easy application of a convergence criterion... may be however that it covers up an insidious trap ...
From...
http://en.wikipedia.org/wiki/Convergent_series
Root test: suppose that the terms of the sequence are non negative and that there exists r such that...
$\displaystyle \lim_{n \rightarrow \infty} \sqrt[n] {a_{n}} = \frac{1}{r}$ (1)
If $\displaystyle r>1$ the series converges, if $\displaystyle r<1$ the series diverges, if $\displaystyle r=1$ the root test is inconclusive...
It is not quite obvious the reason to have in (1) $\displaystyle \frac{1}{r}$ instead of $\displaystyle r$. Anyway in our case is...
$\displaystyle \lim_{n \rightarrow \infty} \sqrt[n] {a_{n}} = \lim_{n \rightarrow \infty} \tan^{-1} \frac{1}{n} = 0$ (2)
Now, if we want to apply the root test as explained by Wiky, the problem is that an $\displaystyle r$ such that $\displaystyle \frac{1}{r}=0$ doesn't exists ...
Merry Christmas from Italy