# Harmonic numbers proof

• Oct 11th 2010, 08:39 AM
Danneedshelp
Harmonic numbers proof
Q: Prove $\displaystyle H_{1}+H_{2}+...+H_{n}=(n+1)H_{n}-n$.

A: First of all, this is only true for $\displaystyle n>1$. So, I let $\displaystyle n=2$ for my base case. The equation held, so I moved on.

Now, I assume $\displaystyle H_{1}+H_{2}+...+H_{k}=(k+1)H_{k}-k$ for some natrual number $\displaystyle k$.

Next, I have that

$\displaystyle H_{1}+H_{2}+...+H_{k}+H_{k+1}=((k+1)H_{k}-k)+H_{k+1}$ by our inductive hypothesis.

I expanded everything on the RHS to get

$\displaystyle (k+1)(1+\frac{1}{2}+...+\frac{1}{k})-k+(1+\frac{1}{2}+...+\frac{1}{k}+\frac{1}{k+1})$
$\displaystyle (k+1)+\frac{k+1}{2}+...+\frac{k+1}{k}-k+(1+\frac{1}{2}+...+\frac{1}{k}+\frac{1}{k+1})$
$\displaystyle (k+2)+\frac{k+2}{2}+...+\frac{k+2}{k}-k+\frac{1}{K+1}$
$\displaystyle (k+2)(1+\frac{1}{2}+...+\frac{1}{k})-k+\frac{1}{k+1}$

Now, I am not sure what to do. I think I may be taking the wrong approach to the problem. Any help would be great.

Thanks
• Oct 11th 2010, 09:01 AM
Traveller
$\displaystyle (k+1)H_k - k + H_{k+1} = (k+1)(H_{k+1} - \frac{1}{k+1}) - k + H_{k+1}$

Can you complete the proof ?