A proof of $\sum_{i=1}^{n}(-1)^{i-1}\binom{n}{i}\frac{1}{i}=\frac{1}{1}+\frac{1}{2}+....+\frac{1}{n}$ 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.
 A: 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.
A: 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 $H_{n+1}$.
