<s>[INST] 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
$
 [/INST] Let us use
$$H_k=- \sum_{m=1}^{k}(-1)^k\frac {{k \choose m}}{m},~~ {n \choose r}={n-1 \choose r-1}+{n-1 \choose r},~~ \sum_{k=m}^{n} {k \choose m}= {n+1 \choose m+1}.$$ Then
$$S_n=\sum_{k=1}^{n} H_k=- \sum_{k=1}^{n} \sum_{m=1}^{n} (-1)^m \frac{{n \choose m}}{m}= -\sum_{m=1}^{n} (-1)^{m} \frac{{n+1 \choose m+1}}{m}
$$ $$\Rightarrow S_n= -\sum_{m=1}^{n} (-1)^m \frac{{n \choose m}}{m}- \sum_{m=1}^{n}(-1)^m \frac{{n \choose m+1}}{m}=H_n- \sum_{m=1}^{n} (-1)^m \frac{n}{m(m+1)} {n-1 \choose m+1}$$ $$\Rightarrow S_n=H_n-n\sum_{m=0}^{n} (-1)^m \left( \frac{1}{m}-\frac{1}{m+1} \right) {n-1 \choose m}$$
$$\Rightarrow S_n= H_n-n\sum_{m=1}^{n-1} (-1)^m \frac{{n-1 \choose m}}{m}-\sum_{m=1}^{n}(-1)^m  \frac{n}{m+1} {n-1 \choose m}=H_n+nH_{n-1}-\sum_{m=1}^{n} (-1)^m {n \choose m+1}$$
$$ \Rightarrow S_n= H_n+nH_{n-1}+\sum_{p=2}^{n} (-1)^p {n \choose p}=H_n+n(H_n-\frac{1}{n})+1-n=(n+1)H_n-n.$$
</s>