# Thread: Root test for convergence/Divergence

1. ## Root test for convergence/Divergence

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?

2. Originally Posted by jzellt
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?

Well the properties of limits for composition functions, we can notice that the inside limit goes to 0 as we got to infinity, now we are approach the value of the arctan of 0 which is?

Correct me if wrong.

3. Any other advice on this one...

4. yes, brush up your notes from calculus I, since RockHard told ya what you need to do.

5. 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...

$\lim_{n \rightarrow \infty} \sqrt[n] {a_{n}} = \frac{1}{r}$ (1)

If $r>1$ the series converges, if $r<1$ the series diverges, if $r=1$ the root test is inconclusive...

It is not quite obvious the reason to have in (1) $\frac{1}{r}$ instead of $r$. Anyway in our case is...

$\lim_{n \rightarrow \infty} \sqrt[n] {a_{n}} = \lim_{n \rightarrow \infty} \tan^{-1} \frac{1}{n} = 0$ (2)

Now, if we want to apply the root test as explained by Wiky, the problem is that an $r$ such that $\frac{1}{r}=0$ doesn't exists ...

Merry Christmas from Italy

6. Well...I'd find wolfram to be more accurate than wiki. Wolfram states this converges by root test, if I entered the equation right