Comments on Profile Post by Sorunome

for which z does this infinite series konverge?

Boo, math.

I'm not sure, but it would seem that z=0 suits it.

I wouldn't be so certain as ((1+(n+1)/(n^2))^(n^2)) goes towards infinity and 0^n is zero, and with zero * infinity you always have to do more investigation

0^n=0=cste for any n>0. Even if series tends to infinity, multiplied by 0 it will be 0. Of course, it does not apply if you multiply this by something, which tends to zero. In that case you should make a full study.

In that case there's no need to worry about limits.

hm, right. Too bad we need to find all z for which it converges :p
I guess i'll think about it some more tomorrow ^.^

It seems that 0 is the only solution. I'll check this tomorrow.

|z| < 1/e ?

Have you been using this:
n*ln[(n+1)/n^2+1] ~ 1?

can't use, we haven't even defined logarythm in the lecture

don't worry, i'll figure it out :p

Makak is right. But I have no idea, how can you find it without Taylor expansion of ln(1+u) with lim(u)=0

Well, we know that e^x = sum(x^n / n!)

IDK i'll figure something out :p

I'm a bit curious, but what have you used to demonstrate this?
(You can do it with Taylor, but there must be another way)

I don't know, i was missing that day in the lecture xD

http://i.imgur.com/QBnYezH.jpg