How to find Sum of Sum of Partial Harmonic Series: $\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} $? How do I find the Sum of the Sum of Finitely Many Harmonic Series: $\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} $?
According to maple it is:
$\sum_{b=1}^{m} \sum_{k=1}^{b} \frac{1}{k} =  \left(  \left( m+1 \right) ^{2}-m-4 \right)  \left( \Psi \left( m+2
 \right) +\gamma \right) + \left( - \left( m+1 \right) ^{2}+2\,m+5
 \right)  \left( \Psi \left( m+3 \right) +\gamma \right) -3
$

The iterated sum actually evaluates to
$$\sum_{b = 1}^m \sum_{k = 1}^b \frac{1}k = \left(m + 1\right)H_m - m.$$
You can show by regrouping terms that for a general sum
$$\sum_{a = 1}^x \sum_{b = 1}^a f(b) = \sum_{k = 1}^x \left(x - k + 1\right) f(k).$$
So plugging in $f(b) = \frac{1}b$, we obtain
\begin{align}
\sum_{a = 1}^x \sum_{b = 1}^a f(b) &= \sum_{a = 1}^x \sum_{b = 1}^a \frac{1}b \\
&= \sum_{k = 1}^x \left(x - k + 1\right) \frac{1}k \\
&= \sum_{k = 1}^x \left(\frac{x}k - 1 + \frac{1}k\right) \\
&= \left(x + 1\right)H_x - x.
\end{align}
I'm guessing that Maple just used a weird polygamma identity somewhere.
