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?
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$\ds{\sum_{i\ =\ 1}^{n}H_{i} = \pars{n + 1}H_{n} - n:\ {\large ?}.\quad
     n\ \geq\ 1}$.
By Induction:
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}
In addittion:

The direct proof becomes:
  \begin{align}&\color{#66f}{\large\sum_{i\ =\ 1}^{n}H_{i}}
=\sum_{i\ =\ 1}^{n}\sum_{k\ =\ 1}^{i}{1 \over k}
=\sum_{k\ =\ 1}^{n}{1 \over k}\sum_{i\ =\ k}^{n}1
=\sum_{k\ =\ 1}^{n}{1 \over k}\pars{n - k + 1}
=\pars{n + 1}\sum_{k\ =\ 1}^{n}{1 \over k} - \sum_{k\ =\ 1}^{n}1
\\[5mm]&=\color{#66f}{\large\pars{n + 1}H_{n} - n}
\end{align}

