Calculate $$\sum_{i=0}^{n-1}H_i\cdot i$$ Where $$H_n = 1+\frac{1}{2}+...+\frac{1}{n} $$
My try
I am interested in using the formula from my lecture: $$ \sum_{i=a}^{b}f(k)\cdot\Delta g(k) = f(b)g(b)-f(a)g(a)-\sum_{i=a}^{b}\Delta f(k)\cdot g(k+1) $$ where $$ \Delta f(k) = f(k+1)-f(k) $$
So in my case I have $$ f(k) = H_k \wedge \Delta g(k) = k $$ so I can take $$ \Delta f(k) = \frac{1}{k+1} \wedge g(k) = \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2} $$
Ok so I can start calculating: $$\sum_{i=0}^{n-1}H_i\cdot i =\\ f(n-1)g(n-1)-0-\sum_{k=0}^{n-1}\frac{1}{k+1} \cdot \frac{k(k+1)}{2} =\\ H_{n-1}\cdot \frac{(n-2)(n-1)}{2}-\frac{1}{2}\cdot\frac{(n-1)n}{2} = \\ \frac{n-1}{2}\cdot \left( H_{n-1}(n-2)-\frac{n}{2} \right) \\ \frac{n-1}{2}\cdot \left( (H_n - \frac{1}{n})(n-2)-\frac{n}{2} \right)$$
But the answer in book is $$ \frac{n(n-1)(2H_n-1)}{4} $$ which is not equal to my result
PS: I can't use $ \int $
By my calculation, the surface term (sorry, physics lingo) should actually be $f\left(b+1\right)g\left(b+1\right)-f\left(a\right)g\left(a\right)$. Cancelling the red terms below, $$\sum_{i=a}^{b}\left(f\left(k\right)\Delta g\left(k\right)+g\left(k+1\right)\Delta f\left(k\right)\right)\\=\sum_{i=a}^{b}\left(\color{red}{f\left(k\right)g\left(k+1\right)}-f\left(k\right)g\left(k\right)+g\left(k+1\right)f\left(k+1\right)\color{red}{-g\left(k+1\right)f\left(k\right)}\right)\\=f\left(b+1\right)g\left(b+1\right)-f\left(a\right)g\left(a\right).$$So instead we get $\frac{n(n-1)}{4}(2H_n-1)$ as required.