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How should I evaluate the sum: $\sum_{k=1}^{n}\frac{1}{k}\binom{n}{k}$ ?

I have no relevant observations or partial results so far.

Any kind of help or advice would be truly appreciated!

Zuzana
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1 Answers1

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Use binom of Newton:

$$(1+x)^n=\sum_{k=0}^n \binom nk x^k\stackrel{\text{integ.}}\implies\frac{(1+x)^{n+1}}{n+1}=\sum_{k=0}^n\binom nk\frac{x^{k+1}}{k+1}+C\;,\;\;C=\text{ constant}$$

Now susbtitute $\;x=0\;$ to find the constant, and then $\;x=1\;$ and do some little algebra.

DonAntonio
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    Please show the algebra to get from $\sum\frac1{k+1}\binom nk$ to $\sum\frac1k\binom nk.$ – bof Mar 29 '16 at 21:57
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    Sorry but is there any relation at all between $\sum\frac1k{n\choose k}$ and $\sum\frac1{k+1}{n\choose k}$? I see none. (Repeating @bof's point.) – Did Mar 29 '16 at 22:11
  • @bof Thanks for pointing the typo, edited. . – DonAntonio Mar 29 '16 at 22:13
  • @did Integrating $\frac{(1+x)^n-1}{x}$ instead should work. – Arthur Mar 29 '16 at 22:18
  • @Arthur Yes, I already figured out that the sum is equal to $\int_0^1\frac{(1+x)^n-1}xdx.$ Can you integrate that in closed form? – bof Mar 29 '16 at 22:23
  • @bof When I stop to think about it, not as anything other than the sum we want to evaluate, no. – Arthur Mar 29 '16 at 22:29
  • @SubhadeepDey The answer is clearly not $1$ for $n\geq 2$. – Arthur Mar 29 '16 at 22:30
  • $$\int_0^1\frac{(1+x)^n-1}xdx=\int_0^1\sum_{k=1}^n\binom nk x^{k-1};dx=\left.\sum_{k=1}^n\binom nk\frac1k\cdot x^k\right|0^1=\sum{k=1}^n\binom nk\frac1k$$ – DonAntonio Mar 29 '16 at 22:30
  • @Arthur Is $\sum_{k=1}^n\frac{2^k-1}k$ an improvement over the sum we want to evaluate? – bof Mar 29 '16 at 22:31
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    @Joanpemo That's right, the integral of $\frac{(x+1)^n-1)}x$ will give the correct answer. Are you capable of calculating that integral? – Arthur Mar 29 '16 at 22:34
  • @Arthur Thank you. Didn't I calculate that integral already in my last comment? Or what do you mean, please? – DonAntonio Mar 29 '16 at 22:35
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    Your last comment gave the RHS, which we can't calculate, so we were resorting to that integral to help us – Nikunj Mar 29 '16 at 22:36
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    I substituted $u=x+1$ and got $$\int_1^2\frac{u^n-1}{u-1}du = \int_1^2(1+u+u^2+\cdots+u^{n-1})du = \sum_{k=1}^n\frac{2^k-1}k.$$ – bof Mar 29 '16 at 22:40