Binomial coefficient challenge

[lfdahl]

Prove the following identity:\[\sum_{n =1}^{\infty }\frac{1}{\binom{n+r}{r+1}}=\frac{r+1}{r},\: \: \: \: r,n \in \mathbb{N}.\]

 

[lfdahl]

Hint:

Spoiler

Consider the difference:

\[\frac{1}{\binom{n+r-1}{r}}-\frac{1}{\binom{n+r}{r}}\]

 

[lfdahl]

Suggested solution:

Spoiler

We have:

\[\frac{1}{\binom{n+r-1}{r}}-\frac{1}{\binom{n+r}{r}}=\frac{r!(n-1)!}{(n+r-1)!}-\frac{r!n!}{(n+r)!} \\\\ =r!(n-1)!\left ( \frac{n+r}{(n+r)!}-\frac{n}{(n+r)!}\right ) = \frac{r!(n-1)!r}{(n+r)!}\]

Now, compare the last term with:

\[\frac{1}{\binom{n+r}{r+1}}=\frac{(r+1)!(n-1)!}{(n+r)!}\]

If we divide by $r$ and multiply by $r+1$, we have the identity:

\[\frac{1}{\binom{n+r}{r+1}}=\frac{r+1}{r}\left ( \frac{1}{\binom{n+r-1}{r}}-\frac{1}{\binom{n+r}{r}} \right )\]

Summing over all possible $n$ (the RHS is a telescoping sum):

\[\sum_{n=1}^{\infty }\frac{1}{\binom{n+r}{r+1}}=\frac{r+1}{r}\sum_{n=1}^{\infty }\left ( \frac{1}{\binom{n+r-1}{r}}-\frac{1}{\binom{n+r}{r}} \right ) \\\\ =\frac{r+1}{r}\left ( \frac{1}{\binom{r}{r}}-\frac{1}{\binom{r+1}{r}}+\frac{1}{\binom{r+1}{r}} -\frac{1}{\binom{r+2}{r}} + \frac{1}{\binom{r+2}{r}} - ... \right ) \\\\ =\frac{r+1}{r}.\]