#what convergence test would I use here?
frigid swift
hardy juncoBOT
steep tide
$\sum_{n=0}^{\infty} \frac{n^n}{2^{n^2}}$
sly schoonerBOT
Tittom_123
steep tide
@frigid swift ?
frigid swift
thx
I'm confused on what convergence test I would use, because I tried the ratio test but the ratio came out to something I couldn't really factor out easily
open oracle
https://cdn.discordapp.com/attachments/1227090522464587836/1227090711514710026/7...
is that supposed to be n = 0 instead of n = 1?
assuming its n = 1, you can do:
,,\sum_{n=1}^\infty\frac{n^n}{2^{n^2}}=\sum_{n=1}^\infty\qty(\frac n{2^n})^n
sly schoonerBOT
mtt
open oracle
which would indicate to use the nth root test
that way, you can figure out whether series converges or not based on $\lim_{n\to\infty}\frac n{2^n}$
sly schoonerBOT
mtt
open oracle
yea it's n=0 my mistake
thats not what I said?
if its n=0 you have 0^0 in the numerator
open oracle
??????
youre giving me two different stories here
anyways usually for questions like this, you ignore what the n = says at the bottom
because you can set that number to be any number you want ahead
vagrant canopy
use alembert