Show that for $k>0$, the series $\sum_{n=1}^{\infty } \binom{k}{n} $ converges absolutely
It is clearly that as $n\longrightarrow \infty $, it will be greater than k eventually. That's my problem, I don't know anything about the combination function with the lower variable bigger than the upper. In order to solve this problem, is there any similar subject or question around here? Could someone point out for me please? Thanks a lot!
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If $k$ is an integer, $\binom kn=0$ for every $n>k$ so the series is a finite sum.
Assume now $N<k<N+1$ for some integer $N$ and let $A:=k(k-1)\dots(k-N).$ Then, for every $n>N,$ $$\left|\binom kn\right|=A\,\frac{(N+1-k)(N+2-k)\dots(n-1-k)}{n!}=B\,\frac{\Gamma(n-k)}{n!}$$ with $B:=\frac A{\Gamma(N+1-k)}.$ By Stirling's formulae, as $n\to\infty,$ $$\left|\binom kn\right|\sim B\frac{\sqrt{\frac{2\pi}{n-k}}\left(\frac{n-k}e\right)^{n-k}}{\sqrt{2\pi n}\left(\frac ne\right)^n}\sim\frac Bn\left(\frac ne\right)^{-k}\left(1-\frac kn\right)^n\sim\frac B{n^{1+k}},$$ whence the convergence, since $1+k>1.$
For $n\in\mathbb{N}$ and $n>k>0$, $$\begin{align*} \left|\frac{a_n}{a_{n+1}}\right|&=\left|\frac{\binom{k}{n}}{\binom{k}{n+1}}\right| \\ &=\left|\frac{k!}{n!(k-n)!}\frac{(n+1)!(k-n-1)!}{k!}\right| \\ &=\left|\frac{(n+1)!}{n!}\frac{(k-n-1)!}{(k-n)!}\right| \\ &=\left|\frac{n+1}{k-n}\right| \\ &=\frac{n+1}{n-k} \end{align*}$$ then by Raabe's test, $$\begin{align*} \lim_{n\to+\infty}\left[n\left(\frac{a_n}{a_{n+1}}-1\right)\right]&=\lim_{n\to+\infty}\left[n\left(\frac{n+1}{n-k}-1\right)\right] \\ &=\lim_{n\to+\infty}\left[\frac{n(1+k)}{n-k}\right] \\ &=1+k>1 \end{align*}$$ the series is absolutely convergent.