determine if this sequence converges absolutely, conditionally or diverge?

(1)^n * n/n^1+1

i know its an alternating series

Question

determine if this sequence converges absolutely, conditionally or diverge?

(1)^n * n/n^1+1

i know its an alternating series

megannicole51 · Answer

@hartnn

agent0smith · Answer

i can't tell what it is...?$$\large \frac{ (-1)^n  n }{ n^2+1 }$$

anonymous · Answer

seems unlikely it says $\frac{(1)^nn}{n^1+1}$

agent0smith · Answer

I just assumed the 1 exponent was a typo

anonymous · Answer

i meant it seems unlikely that that is what the question is
maybe 
$$\frac{(-1)^nn}{n+1}$$ or what you wrote  
$$\frac{(-1)^nn}{n^2+1}$$

megannicole51 · Answer

its n^2 it was a typo...brb

anonymous · Answer

as it alternates, all you have to check is that 
$$\lim_{n\to \infty}\frac{n}{n^2+1}=0$$ which it is

megannicole51 · Answer

simplified to 1/n which diverges

agent0smith · Answer

No but it's the terms that converge to zero (1/n) AND the fact it's alternating, that makes it convergent.

Without the (-1)^n it would be divergent.

agent0smith · Answer

Think of it like this... (basic example)

0.9+0.8+0.7+0.6+0.5+0.4+0.3+0.2+0.1 = 4.5
but
0.9-0.8+0.7-0.6+0.5-0.4+0.3-0.2+0.1 = 0.5

agent0smith · Answer

the alternating adding/subtracting is what makes it converge.

agent0smith · Answer

eg 1/n vs (-1)^n/n: 1/n diverges http://www.wolframalpha.com/input/?i=sum+of+1%2Fn+ (-1)^n/n converges http://www.wolframalpha.com/input/?i=sum+of+%28-1%29%5En%2Fn+