Find the limit of $\frac{n^\alpha}{(1+p)^n}$ as $n$ tends to infinity. $p>0$ and $\alpha$ is real.
I have the solution to this problem but i dont understand the logic behind the steps.
Can someone please help.
EDIT:- Actually i wanted to solve it using the fundamental methods of sequence and series. It begins by saying (1+p)^n is greater than nCk p^k which is greater than n^k* p^k/(2^k*k!)
hence, 0 < $\frac{n^\alpha}{(1+p)^n}$ < (2^k*k!/p^k) n^(alpha-k).
EDIT2 (@Did): It appears that the question is actually to prove that $$(1+p)^n\geqslant{n\choose k} p^k \geqslant n^kp^k/(2^kk!).$$
Edit (3) :- Here's the question and its answer from Rudin's book:
Best Answer
You proved that, for every positive integer $k$ and for every $n\geqslant k$ (actually, for $n\geqslant2k$, see below), $$0 <\frac{n^\alpha}{(1+p)^n} < \frac{2^k\,k!}{p^k}\frac1{n^{k-\alpha}}.$$ For $k\gt\max(1,\alpha)$, this implies that the limit is zero.
Edit: To prove the upper bound above, note that $$(1+p)^n=\sum_{i=0}^n{n\choose i}p^i\geqslant{n\choose k}p^k,$$ and that, for every $n\geqslant2k$, $${n\choose k}=\frac1{k!}\prod_{i=0}^{k-1}(n-i)\geqslant\frac1{k!}\prod_{i=0}^{k-1}\left(\frac{n}2\right)=\frac1{k!}\frac{n^k}{2^k}.$$ Thus, the book probably explains that their upper bound is valid for every $n\geqslant2k$.