The $n^{th}$ Harmonic Number is defined as the summation of the reciprocals of all numbers from $1$ to $n$.
$$H_n = \sum_{i = 1}^n \frac1 i = \frac1 1 + \frac1 2 + \frac1 3 + \frac1 4 + \dots + \frac1 n$$
Lets call the value of $\sum_{j = 1}^n H_j$ as $S_n$
Then,
$\begin{align}
S_n &= H_1 + H_2 + H_3 + \dots + H_n\\[1em]
&= \small \overbrace{\left ( \color{red}{\frac1 1} \right )}^{H_1}
+ \underbrace{\left (\color{red}{\frac1 1} + \color{blue}{\frac1 2} \right )}_{H_2}
+ \overbrace{\left (\color{red}{\frac1 1} + \color{blue}{\frac1 2} + \color{green}{\frac1 3} \right )}^{H_3}
+ \dots
+ \underbrace{\left (\color{red}{\frac1 1} + \color{blue}{\frac1 2} + \color{green}{\frac1 3} + \dots + \frac1 n \right )}_{H_n}\\[1em]
&=\small \color{red}{n \times\frac1 1}+ \color{blue}{ (n-1) \times\frac1 2} + \color{green}{(n-2) \times \frac1 3} + \dots + 1 \times \frac1 n\\[1em]
&= \sum_{i = 1}^n \left (n - i + 1 \right ) \times \frac1 i\\[1em]
&= \sum_{i = 1}^n \left ( \frac{n + 1}{i} - 1 \right )\\[1em]
&= \left ( \sum_{i = 1}^n \frac{\color{red}{n+1}}{i}\right ) - \color{blue}{\left ( \sum_{i = 1}^n 1\right )}\\[1em]
&= \left (\color{red}{(n+1)} \times \underbrace{\sum_{i = 1}^n \frac1 i}_{=H_n}\;\right ) - \color{blue}{n}\\[3em]
\hline
\large S_n &= \large (n+1)\cdot H_n - n
\end{align}$
Hence, the answer is option (B).
