Does this serie converges or diverges?

S=((-1)^n)(n^(1/n))sin(1/n)

n from 1 to infinity

Question

Does this serie converges or diverges?

S=((-1)^n)(n^(1/n))sin(1/n)

n from 1 to infinity

anonymous · Answer

$$\sum_{1}^{\infty} (-1)^n \sqrt[n]{n} \sin(1/n)$$

anonymous · Answer

Who's there  c:

anonymous · Answer

??

experimentx · Answer

i think that will not converge.
for alternating series
lim n->inf |f(n)| = 0

experimentx · Answer

for this particular case, lin n->inf = indeterminate

anonymous · Answer

$$\lim_{n \rightarrow \infty} \sqrt[n]{n} \sin(1/n)=0?? $$

anonymous · Answer

Oh ok

anonymous · Answer

So , how can I prove if the limit is indeterminate?

experimentx · Answer

experimentx · Answer

oh ... sorry. my mistake.

experimentx · Answer

i thought n->0 inside sin
it's zero.

anonymous · Answer

ok, so limit is 0 right?

experimentx · Answer

well ... one condition is satisfied. the other condition is 
|a_n| > |a_n+1|

anonymous · Answer

Yep

experimentx · Answer

i guess this means it does not converge. http://www.wolframalpha.com/input/?i=Sum [%28-1%29^n+n^%281%2Fn%29+*+sin%281%2Fn%29%2C+1%2C+infinity] n^n sin(1/n) > (n+1)^(n+1) sin(1/n+1)

anonymous · Answer

can you copy paste the link again?

zarkon · Answer

it converges

anonymous · Answer

Could you explain it?

zarkon · Answer

I could

experimentx · Answer

zarkon · Answer

show that $\sin(1/n)$ decreases for all $n$

and show that $n^{1/n}$ decreases for all $n\ge 3$

experimentx · Answer

zarkon · Answer

$$n^{1/n}$$ is not real for most negative values of n\]

anonymous · Answer

Ok so as limit -->0 and it's a decreasing sequence then the series converges by Leibniz test right?

zarkon · Answer

yes...the alternating series test

anonymous · Answer

Ok, thank you guys!

experimentx · Answer