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I want to compute this expression without using a ' known induction to apply':

$$\sum_{k=0}^{n}{\binom{n}{k}\frac{1}{2k+1}}$$

I saw this one :Is the numerator of $\sum_{k=0}^{n}{(-1)^k\binom{n}{k}\frac{1}{2k+1}}$ always a power of $2$ in lowest terms?

First proof is by induction, the second one uses hypergeometric function which is a bit to non-elementary.

Thanks.

(I tried Pascal, then some identity on binomial coeff , the sum I want to calculate is from the associated integral from zero to one)

ROULEAU PQ
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  • Seeing your user name, may I bet that you are French ? Just out of curiosity : where are you located ? I am in Pau. – Claude Leibovici May 27 '18 at 13:28
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    Consider $(x^2+1)^n=\sum_{n=1}^{\infty}\binom{n}{k}x^{2k}$, then $\int{}(x^2+1)^ndx$ is $\sum_{n=1}^{\infty}\binom{n}{k}\frac{1}{2k+1}x^{2k+1}$. Can you progress somehow from here? – Μάρκος Καραμέρης May 27 '18 at 13:32
  • Thanks for answering, I started from the integral to go to the sum, but you can see it just as $ \int_0^1 (x^2+1)^n dx $ . I tought about kind of Cesaro summation (because the present sum is in fact an average , weighted by binomial coefficient of whom the sum gives $2^n$ . – ROULEAU PQ May 27 '18 at 19:14
  • Coming from Rhônes-Alpes . – ROULEAU PQ May 27 '18 at 19:21
  • Coming from Rhônes-Alpes .Does Pau has a well preserved old-city ? – ROULEAU PQ May 27 '18 at 19:27

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