Convergence of series - Which test? I want to check for convergence the series $\displaystyle{\sum_{n=m}^{\infty}\binom{n}{m}^{-1}}$. 
$$$$ 
I have done the following: 
We use the ratio test. 
We have that $$a_n=\binom{n}{m}^{-1}=\left (\frac{n!}{m!(n-m)!}\right )^{-1}=\frac{m!(n-m)!}{n!}$$ and $$a_{n+1}=\binom{n+1}{m}^{-1}=\left (\frac{(n+1)!}{m!(n+1-m)!}\right )^{-1}=\frac{m!(n+1-m)!}{(n+1)!}$$ 
Therefore we get \begin{align*}\left |\frac{a_{n+1}}{a_n}\right |&=\frac{m!(n-m+1)!}{(n+1)!}\cdot \frac{n!}{m!(n-m)!}\\ & =\frac{(n-m)!(n-m+1)}{n!(n+1)}\cdot \frac{n!}{(n-m)!}\\ & =\frac{(n-m+1)}{(n+1)}\underset{n\rightarrow \infty}{\longrightarrow }1\end{align*} 
So, we cannot say something by the ratio test. Which convergence test should we use instead? 
 A: Better than any test, there is an explicit computation. For any $m\geq 2$,
$$\sum_{n\geq m}\binom{n}{m}^{-1} = \sum_{n\geq m}\frac{(n-m)!m!}{n!}=m\sum_{n\geq m}\frac{\Gamma(n-m+1)\Gamma(m)}{\Gamma(n+1)} $$
can be written as
$$ m\sum_{n\geq m}B(m,n-m+1) = m\sum_{n\geq m}\int_{0}^{1}x^{m-1}(1-x)^{n-m}\,dx $$
or as
$$ m\int_{0}^{1}x^{m-1}\sum_{n\geq m}(1-x)^{n-m}\,dx=m\int_{0}^{1}x^{m-2}\,dx=\color{red}{\frac{m}{m-1}}. $$
A: This is a comment on
Jack D'Aurizio's answer.
He shows that
$\sum_{n\geq m}\binom{n}{m}^{-1} 
=\dfrac{m}{m-1}
$.
Expanding the two sides
in power series,
$\dfrac{m}{m-1}
=\dfrac{1}{1-1/m}
=\sum_{n=0}^{\infty} \dfrac1{m^n}
=\sum_{n=0}^{\infty} \dfrac1{\prod_{k=1}^n m}
$
and
$\begin{array}\\
\sum_{n\geq m}\binom{n}{m}^{-1} 
&= \sum_{n\geq m}\dfrac{(n-m)!m!}{n!}\\
&= \sum_{n\geq 0}\frac{(n)!m!}{(n+m)!}\\
&= \sum_{n=0}^{\infty}\dfrac{n!m!}{m!\prod_{k=m+1}^{n+m}k}\\
&= \sum_{n=0}^{\infty}\dfrac{n!}{\prod_{k=m+1}^{n+m}k}\\
&= \sum_{n=0}^{\infty}\dfrac{\prod_{k=1}^n k}{\prod_{k=1}^{n}(k+m)}\\
&= \sum_{n=0}^{\infty}\dfrac{1}{\prod_{k=1}^{n}(1+m/k)}\\
\text{so}\\
\sum_{n=0}^{\infty} \dfrac1{\prod_{k=1}^n m}
&= \sum_{n=0}^{\infty}\dfrac{1}{\prod_{k=1}^{n}(1+m/k)}\\
\end{array}
$
I find this curious.
A: Use  equivalents:
For $m$ fixed, we have $\dbinom nm=\dfrac{n(n-1)\dots(n-m+1)}{m!}\sim_\infty\dfrac{n^m}{m!}$, so
$$\dbinom nm^{\mkern-4mu-1}\sim_\infty\frac{m!}{n^m},$$
and the latter converges if and only if $m\ge 2$.
A: If $m=0$, the terms do not go to $0$, so the series diverges.
If $m=1$, this is the harmonic series, which diverges.
If $m\ge2$,
$$
\begin{align}
\sum_{n=m}^\infty\binom{n}{m}^{-1}
&=\sum_{n=m}^\infty\frac{m!}{n(n-1)\cdots(n-m+1)}\tag1\\
&\le\sum_{n=m}^\infty\frac{m!}{(n-m+1)^m}\tag2\\
&=\sum_{n=1}^\infty\frac{m!}{n^m}\tag3
\end{align}
$$
Explanation:
$(1)$ rewrite the binomial coefficient
$(2)$ replace each factor in the denominator by the smallest factor
$(3)$ substitute $n\mapsto n+m-1$
and $(3)$ converges, so the original series converges by comparison.
