Ways to show $\left (\frac{n^2}{2^n} \right)$ is converging to zero

Question

I still have problems understanding how to prove that a sequence converges, stuck with

$a_n=\left(\frac{n^2}{2^n} \right)$ I know that $\lim_{n\to\infty} a_n=0$ but don't know how to proceed. I tried replacing $2^n$ with $\sum_{k = 0}^n { \binom{n}{k}}$ to get $\left(\frac{n^2}{\sum_{k = 0}^n \left(\frac{n!}{k!(n-k)!} \right)} \right)$. Next I want to manipulate the term somehow (don't understand how this could be done) to get $n$ instead of $n^2$ as numerator to show that the denominator is always bigger than $n^2$ starting from some $n_0$. How would I then get the value of $\epsilon$?

Is this a valid approach? How can I extrace a $n$ from the sum? Also different approaches are appriciated!

@edit forgot to include that I want to show that this sequence converges to $0$, so showing convergence is not enough. In the linked dublicate it's written as fact if $a_n > 0$ and the ratio test $< 1$. Why is this true?

And a more general question https://math.stackexchange.com/questions/2489665/prove-that-if-k-in-mathbb-n-and-a1-then-lim-limits-n-to-infty-fra/ — rtybase, Jan 28 '18 at 13:05

score 1 · Answer 1 · answered Jan 28 '18 at 13:24

1

by using the ratio test $$\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_n} \right |<1$$ $$\lim_{n\rightarrow \infty}\frac{(n+1)^2}{2^{n+1}}\frac{2^n}{n^2}=\lim_{n\rightarrow \infty}\frac{1}{2}(1+\frac{1}{n})^2=\frac{1}{2}<1$$ so the sequence converges

answered Jan 28 '18 at 13:24

E.H.E

23,280

this shows convergence, how can i show that it converges to zero? – JDizzle Jan 28 '18 at 13:55
@JDizzle Do you ask about series or sequence?? – E.H.E Jan 28 '18 at 14:03
I ment sequence, the series was used to replace $2^n$. – JDizzle Jan 28 '18 at 14:08

Ways to show $\left (\frac{n^2}{2^n} \right)$ is converging to zero

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