$\sum_{k=1}^nH_k = (n+1)H_n-n$. Why? This is motivated by my answer to this question.  
The Wikipedia entry on harmonic numbers gives the following identity: 
$$
\sum_{k=1}^nH_k=(n+1)H_n-n
$$
Why is this?  
Note that I don't just want a proof of this fact (It's very easily done by induction, for example).  Instead, I want to know if anyone's got a really nice interpretation of this result: a very simple way to show not just that this relation is true, but why it is true.  
Has anyone got a way of showing that this identity is not just true, but obvious?
 A: Consider the different terms $\frac 1k$. Once you added the $n$ terms $\,H_k\,$ you'll have :
\begin{array} {lcc}
&n &\text{terms} &1\\
&n-1&\text{terms}&\frac 12\\
&\cdots\\
&n+1-k&\text{terms}&\frac 1k\\
&\cdots\\
&2&\text{terms} &\frac 1{n-1}\\
&1&\text{term}&\frac 1n\\
\end{array}
The sum will be $$\sum_{k=1}^nH_k=\sum_{k=1}^n \frac{n+1-k}k=(n+1)H_n-n$$
A: $$
\begin{align}
\sum_{k=1}^nH_k
&=\sum_{k=1}^n\sum_{j=1}^k\frac1j&&\text{expand }H_k\\
&=\sum_{j=1}^n\sum_{k=j}^n\frac1j&&\begin{array}{}\text{change order of summation}\\\text{where }1\le j\le k\le n\end{array}\\
&=\sum_{j=1}^n\frac{n-j+1}{j}&&\begin{array}{}\text{each term is constant over}\\\text{the inner summation}\end{array}\\
&=(n+1)\sum_{j=1}^n\frac1j-\sum_{j=1}^n1&&\text{separate terms}\\[6pt]
&=(n+1)H_n-n&&\text{sum}
\end{align}
$$
A: I would see that as the discrete version of $\int_1^x\ln t \,dt= x\ln x-x+1$. This leads you to a proof by Abel transformation (discrete integration by parts):
$$
\sum_{k=1}^nH_k=\sum_{k=1}^nH_k(k+1-k)=(n+1)H_{n+1}-1H_1-\sum_{k=1}^n\underbrace{(k+1)(H_{k+1}-H_k)}_{=1}
$$
$$
=(n+1)H_{n+1}-1-n=(n+1)H_n-n.
$$
A: Let $a_1, a_2,,\ldots$ be any sequence, let $H_n$ denote the partial sum 
$$H_n=a_1+a_2+\cdots+a_n$$
and let $K_n$ denote the partial sum
$$K_n=1a_1+2a_2+\cdots+na_n$$
Then 
$$\begin{align}
\sum_{k=1}^n H_k &= a_1+(a_1+a_2)+\cdots(a_1+a_2+\cdots+a_n)\\
&=na_1+(n-1)a_2+\cdots+2a_{n-1}+1a_n\\
&=(n+1)(a_1+a_2+\cdots a_n)-(1a_1+2a_2+\cdots+na_n)\\
&=(n+1)H_n - (1a_1+2a_2+\cdots+na_n)\\
&=(n+1)H_n-K_n\\
\end{align}$$
Given this completely general identity, the specific identity for the harmonic numbers is obvious from the simple observation that
$$K_n={1\over1}+{2\over2}+\cdots+{n\over n}=n$$
A: I suck at making pictures, but I try nevertheless. Write $n+1$ rows of the sum $H_n$:
$$\begin{matrix}
1 & \frac12 & \frac13 & \dotsb & \frac1n\\
\overline{1\Big\vert} & \frac12 & \frac13 & \dotsb & \frac1n\\
1 & \overline{\frac12\Big\vert} & \frac13 & \dotsb & \frac1n\\
1 & \frac12 & \overline{\frac13\Big\vert}\\
\vdots & & &\ddots & \vdots\\
1 & \frac12 &\frac13 & \dotsb & \overline{\frac1n\Big\vert}
\end{matrix}$$
The total sum is obviously $(n+1)H_n$. The part below the diagonal is obviously $\sum\limits_{k=1}^n H_k$. The part above (and including) the diagonal is obviously $\sum_{k=1}^n k\cdot\frac1k = n$.
It boils down of course to the same argument as Raymond Manzoni gave, but maybe the picture makes it even more obvious.
A: Let $S_n$ denotes $\displaystyle\sum_{k=1}^n H_k$ and by applying Abel's summation:
$$\displaystyle\sum_{k=1}^n a_k b_k=A_nb_{n}+\sum_{k=1}^{n-1}A_k\left(b_k-b_{k+1}\right)\ $$ where $\ \displaystyle A_n=\sum_{i=1}^n a_i\ $ and letting $\ \displaystyle a_k=1 $ , $\ \displaystyle b_k=H_k\ $, we get,
\begin{align}
S_n&=\left(\sum_{i=1}^n1\right)H_{n}+\sum_{k=1}^{n-1}\left(\sum_{i=1}^k1\right)\left(H_k-H_{k+1}\right)\\
&=nH_{n}+\sum_{k=1}^{n-1}(k)\left(-\frac1{k+1}\right), \quad\color{blue}{\sum_{k=1}^{n-1}\frac{k}{k+1}=\sum_{k=0}^{n-1}\frac{k}{k+1}}\\
&=nH_{n}-\sum_{k=1}^n\frac{k-1}{k}\\
&=nH_{n}-\sum_{k=1}^n1+\sum_{k=1}^n\frac{1}{k}\\
&=nH_n-n+H_n\\
&=H_n(n+1)-n\\
\end{align}
