Sum of Scaled Harmonic Numbers I came across the following identity:
$$
\frac{1}{n}\sum_{j=1}^{n}\sum_{k=j}^{n}\frac{1}{k}=1.
$$
However, I do not know how to prove it except verify it by numerical calculations.
Does any one know how to prove it or give me some hints? Thanks very much.
 A: Using Iverson Brackets can simplify the changing of the order of summation:
$$
\begin{align}
\frac1n\sum_{j=1}^n\sum_{k=j}^n\frac1k
&=\frac1n\sum_{j=1}^n\sum_{k=1}^n[k\ge j]\frac1k\\
&=\frac1n\sum_{k=1}^n\sum_{j=1}^n[k\ge j]\frac1k\\
&=\frac1n\sum_{k=1}^n\sum_{j=1}^k\frac1k\\
&=\frac1n\sum_{k=1}^n1\\[9pt]
&=1
\end{align}
$$
A: In the double sum
$$\sum_{j=1}^{n}\sum_{k=j}^{n}\frac{1}{k}$$
each fraction $1/k$ occurs $k$ times, so overall these contribute $1$ to the sum.
As $k$ ranges from $1$ to $n$, the total sum is $n$.
A: We can also write the index region conveniently to better see what's going on.

We obtain
  \begin{align*}
\frac{1}{n}\sum_{j=1}^n\sum_{k=j}^n\frac{1}{k}&=\frac{1}{n}\sum_{\color{blue}{1\leq j\leq k\leq n}}\frac{1}{k}\\
&=\frac{1}{n}\sum_{k=1}^n\sum_{j=1}^k\frac{1}{k}\\
&=\frac{1}{n}\sum_{k=1}^n1\\
&=1
\end{align*}

A: Another proof uses the well-known formula for the sum of harmonic numbers
$$\sum_{k=1}^{n} H_k = (n+1)H_n -n$$
The double sum of the OP can be written as
$$\frac{1}{n} \sum_{j=1}^{n}\left (H_n - H_{j-1}\right)\\
=H_n - \frac{1}{n} \sum_{m=1}^{n-1} H_m\\
= H_n - \frac{1}{n} \left (n H_{n-1} -n +1\right)\\
= H_n - H_{n-1} + 1 -\frac{1}{n} = 1$$
In the second line we have used that $H_0=0$, and in last line we have employed the defining recursion relation of the harmonic numbers.
A: By double counting we have
$$\frac{1}{n}\sum_{j=1}^{n}\sum_{k=j}^{n}\frac{1}{k}=\frac1n \sum_{k=1}^{n}\sum_{j=1}^{k}\frac{1}{k}=\frac1n \sum_{k=1}^{n}1=\frac1n \cdot n=1$$

