# Prove that the sum of harmonic series 1..n can be expressed as (n+1)H_n -n

Prove by induction that the sum of harmonic series Hn from 1 to n where n is a natural number is as follows. $$H_n = \sum\limits_{i=1}^n 1/i$$ Prove: $$\sum\limits_{i=1}^nH_i = (n+1)H_n -n$$ n=1 $$H_1 = 1 = (1+1)(H_1) -1 = 2(1)-1 =1$$ n=k+1 $$H_{n+1} = H_n +1/(n+1)$$ $$H_1 +H_2 +... +H_{n+1} = (n+1 +1 )H_{n+1} -(n+1)$$ $$H_1 +H_2 +... +H_{n+1} = (n+2)H_{n+1} -n -1$$ $$H_1 +H_2 +... +H_{n+1} = (n+2)H_{n+1} -n -1$$ $$H_1 +H_2 +... +H_{n+1} = nH_{n+1} +2H_{n+1} -n -1$$ $$H_1 +H_2 +... +H_{n} = nH_{n+1} +2H_{n+1} -n -1 = (n+2)(H_{n+1}) -n -1$$ $$H_1 +H_2 +... +H_{n} =(n+2)(H_n +1)/(n+1) -n -1$$ At this point I get lost. I've been at this for a while and I don't know if I'm even on the right track, does anyone have a solution. Did I go wrong anywhere?

• You only need to write, since $H_1=1$, that $H_1 = 2 H_1-1$, but you really need to include $2H_1-1$ for clarity. – Thomas Andrews Nov 5 '14 at 3:14
• Induction is a good strategy. But don't assume the thing you're trying to prove ... you do that in the second line of your seven-equation group. Instead, start from $H_{n+1} = H_n + \frac1{n+1}$ and plug in the induction hypothesis for $H_n$; see if you can rewrite the resulting expression to deduce that second line (rather than assuming it). – Greg Martin Nov 5 '14 at 3:25

It's obviously true for $\ds{n = 1}$. Lets assume that it's true for a given $\ds{n\ >\ 1}$. Then, \begin{align} &\color{#c00000}{\sum_{i\ =\ 1}^{n + 1}H_{i}} =\sum_{i\ =\ 1}^{n}H_{i} + H_{n + 1} =\pars{n + 1}H_{n} - n + H_{n + 1} \\[5mm]&=\pars{n + 1}\pars{H_{n + 1} - {1 \over n + 1}} - n + H_{n + 1} =\pars{n + 1}H_{n + 1} - 1 - n + H_{n + 1} \\[5mm]&=\color{#c00000}{\bracks{\pars{n + 1} + 1}H_{n + 1} - \pars{n + 1}} \end{align}