# an infinite sum with binomials

I've got a new exercise. $$\sum_{k=0}^{n} \frac{\binom{n}{k}}{n2^n+k}$$ I've tried this: $$(1+x)^n=\sum_{k=0}^n\binom{n}{k}1^{n-k}x^k$$ Substitute $$x=x^{n2^n}$$ And integrate from 0 to 1 with respect to x. $$\int_0^1(1+x^{n2^n})^n = 1+\frac{1}{n2^n+1}\binom{n}{1}+\frac{1}{n2^n+2}\binom{n}{2}+...++\frac{1}{n2^n+n}\binom{n}{n} | -1+\frac{1}{n2^n}$$

$$\int_0^1(1+x^{n2^n})^n -1+\frac{1}{n2^n} = \sum_{k=0}^{n} \frac{\binom{n}{k}}{n2^n+k}$$ And from here i have no idea!Thank you for your time!

• You unfortunately have a wrong implementation of a good idea. For $n \ge 1$ we have \begin{align} \int_0^1 x^{n2^n - 1}(1+x)^n dx &= \int_0^1 \sum_{k=0}^n {n \choose k} x^{n2^n + k - 1} dx = \sum_{k=0}^n {n \choose k} \left. \frac{x^{n2^n + k}}{n2^n + k} \right|^1_0= \sum_{k=0}^n {n \choose k} \frac{1}{n2^n + k} \; . \end{align} The form of the integral is reminescent of the definition of the beta function. – Jordan Payette Jun 4 '18 at 11:04
• If you are interested in the limit of this expression: Limit of sum of terms containing binomial coefficients. – Martin Sleziak Jun 4 '18 at 11:59