$$\sum _{i=1}^n\left(-1\right)^{i+1}\:\binom{n}{i}\:\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$ Can someone give me a hint on how to give the proof, I am stuck when I am proving it for p(n+1):
$$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}+\left(-1\right)^{n+2}\cdot \frac{1}{n+1}$$
is there a handy way to get rid of the $-1$?
$\sum \:_{i=1}^{n+1}\left(-1\right)^{i+1}\:\binom{n+1}{i}\:\frac{1}{i}=\sum \:_{i=1}^{n+1}\left(-1\right)^{i+1}\:\left[\binom{n}{i}+\binom{n}{i-1}\right] \frac{1}{i}=\sum \:\:_{i=1}^{n+1}\left(-1\right)^{i+1}\binom{n}{i}\frac{1}{i}+\binom{n+1}{i} \frac{1}{n}$
HINT :
Use $$\binom{n}{i}=\binom{n-1}{i}+\binom{n-1}{i-1}$$ and $$\binom{n-1}{i-1}\frac 1i=\frac 1n\binom{n}{i}.$$