How can I prove $$\sum_{i=1}^{n}(-1)^{i-1}\dbinom{n}{i}\frac{1}{i}=\frac{1}{1}+\frac{1}{2}+....+\frac{1}{n}\quad ?$$ I have tried induction, but it doesn't work.
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2See http://math.stackexchange.com/questions/437523/proving-binomial-idenity-without-calculus – lab bhattacharjee Sep 20 '15 at 04:50
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You can prove it by using definite integration. See my answer below. – Olivier Oloa Sep 20 '15 at 05:15
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Idea: Call the LHS $\sum_i f(i,n)$ for a moment. Consider $\sum_n\sum_i f(i,n)x^n$, switch the summation, and simplify. At the end of the simplification, find out what the coefficient of $x^n$ is. (Remember that $\sum_n\binom nix^n=\frac{x^i}{(x-1)^{i+1}}$.) – Akiva Weinberger Sep 20 '15 at 05:59
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Hint. You may consider $$ I_n:=\int_0^1\frac{1-x^n}{1-x}dx. $$ On the one hand use
$$ \frac{1-x^{n}}{1-x}=1+x+\cdots+x^{n-1},\qquad x\neq 1, $$
then integrate the finite sum termwise.
On the other hand make the change of variable $t:=1-x$ giving
$$ \int_0^1\frac{1-x^n}{1-x}dx=\int_0^1\frac{1-(1-t)^n}{t}dt $$
then expand the integrand by using the binomial theorem and integrate termwise.
Olivier Oloa
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I tried your second one by expand the binomial theorem and integrate termwise but I cannot get the same as mine. – John Sep 20 '15 at 05:16
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@John You get $$ \frac{1-(1-t)^n}{t}=\sum_{i=1}^{n}(-1)^{i-1}\dbinom{n}{i}t^{i-1}$$ then you just integrate each term using $\int_0^1t^{i-1}dt=\frac{1}{i}$. – Olivier Oloa Sep 20 '15 at 05:19
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Hence shouldn't there be a minus sign in you last integral equality – Sayan Chattopadhyay May 23 '16 at 06:23
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Yes @OliverOla in your last integral inequality where you substituted t=1-x. So $dt=-dx$ – Sayan Chattopadhyay May 23 '16 at 06:51
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I think you are forgetting the bounds: $1 \to 0$ and $0 \to1$, so there is no minus sign to put. – Olivier Oloa May 23 '16 at 06:53
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Induction works fine, though you have to be a little clever at one point. For convenience I’ll use the standard notation
$$H_n=\sum_{i=1}^n\frac1n\;.$$
Now using the fact that $\frac1k\binom{n-1}{k-1}=\frac1n\binom{n}k$, we have
$$\begin{align*} \sum_{i=1}^{n+1}(-1)^{i-1}\binom{n+1}i\frac1i&\overset{(1)}=\sum_{i=1}^{n+1}(-1)^{i-1}\frac1i\left(\binom{n}i+\binom{n}{i-1}\right)\\ &=\sum_{i=1}^{n+1}(-1)^{i-1}\frac1i\binom{n}i+\sum_{i=1}^{n+1}(-1)^{i-1}\frac1i\binom{n}{i-1}\\ &\overset{(2)}=\sum_{i=1}^n(-1)^{i-1}\frac1i\binom{n}i+\sum_{i=1}^{n+1}(-1)^{i-1}\frac1{n+1}\binom{n+1}i\\ &\overset{(3)}=H_n+\frac1{n+1}\sum_{i=1}^{n+1}(-1)^{i-1}\binom{n+1}i\;, \end{align*}$$
where $(1)$ uses Pascal’s identity, $(2)$ uses the identity mentioned above, and $(3)$ uses the induction hypothesis. Thus,
$$\left(1-\frac1{n+1}\right)\sum_{i=1}^{n+1}(-1)^{i-1}\binom{n+1}i=H_n\;,$$
and you can easily solve for $\sum_{i=1}^{n+1}(-1)^{i-1}\frac1i\binom{n+1}i$ to find that it is indeed $H_{n+1}$.
Brian M. Scott
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