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\[\sum_{n=0}^{\inf} \frac{n+3^n}{4^n}\]

I would seperate it in two parts: n/4^n+(3/4)^n
Then I'd try the root test on both parts seperately.

So i would be left with (n/4) = infinity= div
(3/4) <1 = conv..
Because one div, they both dierge???

only the 4 is raised to the power n.

Nope, so is the 3.

in the first part I mean.

So you don't get n/4 when you do the root test.

Then what do i get??

\[\frac{\sqrt[n]{n}}{4}\]

lol, that looks more complicated. How would i solve that??

So \[\frac{\sqrt[n]{n}}{4} \rightarrow \frac{1}{4}\]
So that part does converge.

I do know the n term test actually, thanks wiki.

It's the first one to try, because it's easy to calculate.

general rule: if you see powers of n: use the ratio test,
if you see factorials: ratio test.

Anyway, good luck.

thaaanks!!