How to sum the sequence $$\sum_{k=0}^n \frac{\binom n k}{(k+1)(k+3)} $$
where $n \choose k$ are the usual binomial coefficients in expansion of $(1+x)^n$.
I know how to sum such sequence when the denominator has consecutive multiplicands (by repeatedly integrating the expansion of $(1+x)^n$).
Does there even exist a general procedure for such sums?
Best Answer
One can proceed similarly as in Is there a closed formula for $\sum_{k=0}^{n}\frac{1}{(k+1)(k+2)}\binom{n}{k}$? : Integrating $$ \sum_{k=0}^n \binom n k x^k = (1+x)^n $$ gives $$ \sum_{k=0}^n \binom n k \frac{x^{k+1}}{k+1} = \frac{1}{n+1}\bigl((1+x)^{n+1} - 1 \bigr). $$ Now multiply by $x$ and then integrate again, for $0 \le x \le 1$: $$ \sum_{k=0}^n \binom n k \frac{1}{(k+1)(k+3)} = \frac{1}{n+1} \int_0^1 x\bigl((1+x)^{n+1} - 1 \bigr) \, dx \, . $$ The integral can be evaluated with integration by parts, the result is $$ \sum_{k=0}^n \binom n k \frac{1}{(k+1)(k+3)} = \frac{2^{n+3} - (n+4)}{2(n+2)(n+3)} \, . $$