Mathematical induction with two nested sums

Question

$\sum_{j=1}^{n} {(2j-1)}{\left(\sum_{m=j}^{n} \frac{1}{m}\right)} = \frac{n(n+1)}{2}$

Would anyone show how to prove this by induction?

Answer 1

The simplest way to show this is just to exchange the two sums, but it wouldn't need induction.

If induction is really needed, you can do it this way (the base case is trivial).

Suppose your equality true for a $n\in\mathbb N$.

Then

$$\sum_{j=1}^{n+1} {(2j-1)}{\left(\sum_{m=j}^{n+1} \frac{1}{m}\right)} = \sum_{j=1}^{n} {(2j-1)}{\left(\sum_{m=j}^{n+1} \frac{1}{m}\right)} + (2(n+1) - 1)\times \frac{1}{n+1} $$ $$ = \sum_{j=1}^{n} {(2j-1)}{\left(\sum_{m=j}^{n} \frac{1}{m} + \frac{1}{n+1}\right)} + (2(n+1) - 1)\times \frac{1}{n+1} $$ $$ = \sum_{j=1}^{n} {(2j-1)}{\left(\sum_{m=j}^{n} \frac{1}{m}\right)} + \sum_{j=1}^{n} {(2j-1)}\frac{1}{n+1} + (2(n+1) - 1)\times \frac{1}{n+1} $$ $$ = \frac{n(n+1)}2 + \frac{n^2}{n+1} + 2 - \frac1{n+1}$$ $$ = \frac{(n+1)(n+2)}2$$