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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!

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  • $\begingroup$ 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. $\endgroup$ – Jordan Payette Jun 4 '18 at 11:04
  • $\begingroup$ If you are interested in the limit of this expression: Limit of sum of terms containing binomial coefficients. $\endgroup$ – Martin Sleziak Jun 4 '18 at 11:59

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