Question

Does ∑ 1 to ∞ (n^2 ÷(e^n /3) ) converge or diverge?

Answer 1

You need to apply a test for converge, which one that you know of works here?

Answer 2

I'm thinking Limit Comparision Test but I'm not sure which series would work to compare it with

Answer 3

I tried e^n/3 but that doesn't work

Answer 4

hmm, that's not what I had in mind. It might work if you choose the other series properly, but I'd try a different test.

Answer 5

Which test do you think would work?

Answer 6

the root test.

Answer 7

I know the root test but this problem was in a section in the book before the root test (by then, there were only nth term test, DCT, LCT, Integral Test) so it should be solvable by one of these tests

Answer 8

Ok, the root test would be a lot easier, but the integral test might work.

Answer 9

I haven't thought about solving it using the intergral test really, so I'll try that I guess and see where I get

Answer 10

You need to integrate by parts twice, so have fun.

Answer 11

Yep, that works

Answer 12

\[\sum_{k=1}^{\infty} \frac{k^2}{\frac{e^k}{3}}=3\sum_{k=1}^{\infty}\frac{k^2}{e^k}\]
There exists m (natural number) such that m < k implies k^4 < e^k. Then\[3\sum_{k=1}^{\infty}\frac{k^2}{e^k}=3(\sum_{k=1}^{m}\frac{k^2}{e^k}+\sum_{k=m+1}^{\infty}\frac{k^2}{e^k})\le 3(\sum_{k=1}^{m}\frac{k^2}{e^k}+\sum_{k=1}^{m}\frac{k^2}{k^4})\]
In the expression, the first term is finite (a definite sum), the second converges. Convergence by comparison test.

Answer 13

Limits of summation on the last term should be k=m+1 to infinity. Sorry for the shoddy attention to detail....