Showing $\sum_{i=1}^n\sum_{j=1}^n\frac1{ij}=\sum_{i=1}^n(\sum_{j=1}^i+\sum_{j=i}^n)\frac{1}{ij}-\sum_{i=1}^n\frac1{i^2}$

Question

I was reading Almost Impossible Integrals by Cornel, where I encountered this manipulation

$$\sum_{i=1}^n\sum_{j=1}^n \dfrac{1}{ij}= \sum_{i=1}^n \left(\sum_{j=1}^i+\sum_{j=i}^n\right)\dfrac{1}{ij}-\sum_{i=1}^n\dfrac{1}{i^2}$$

I am new to such manipulations, and didn't understand how this works. I know the basics symmetry of summand, but can't deduce this.

Can you please help me understanding this?

Also, after this step, he writes

$$ \sum_{i=1}^n \left(\sum_{j=1}^i+\sum_{j=i}^n\right)\dfrac{1}{ij} = 2\sum_{i=1}^n \sum_{j=1}^i \dfrac{1}{ij}$$

How does this happen?

you are adding and subtracting the case $j=i$ in the inner sum — Giulio R, Jun 22 '21 at 17:09

CY Aries · Accepted Answer · 2021-06-22T18:44:59.677

2

$j=i$ is used twice.

$\displaystyle \left(\sum_{j=1}^i+\sum_{j=i}^n\right)\frac1{ij}=\left(\sum_{j=1}^i+\sum_{j=i+1}^n\right)\frac1{ij}+\frac1{i^2}=\sum_{j=1}^n\frac1{ij}+\frac1{i^2}$

For the second step, we have

\begin{align*} \sum_{i=1}^n\sum_{j=i}^na_{i,j}&=\sum_{i=1}^n(a_{i,i}+a_{i,i+1}+a_{i,i+2}+\cdots+a_{i,n})\\ &=(a_{1,1}+a_{1,2}+a_{1,3}+\cdots+a_{1,n})\\ &\qquad+(a_{2,2}+a_{2,3}+a_{2,4}+\cdots+a_{2,n})\\ &\qquad+(a_{3,3}+a_{3,4}+a_{3,5}+\cdots+a_{3,n})\\ &\qquad+\cdots\\ &\qquad+(a_{n-1,n-1}+a_{n-1,n})\\ &\qquad+a_{n,n}\\ &=a_{1,1}\\ &\qquad+(a_{1,2}+a_{2,2})\\ &\qquad+(a_{1,3}+a_{2,3}+a_{3,3})\\ &\qquad+\cdots\\ &\qquad+(a_{1,n-1}+a_{2,n-1}+a_{3,n-1}+\cdots+a_{n-1,n-1})\\ &\qquad+(a_{1,n}+a_{2,n}+a_{3,n}+\cdots+a_{n-1,n}+a_{n,n})\\ &=\sum_{j=1}^n(a_{1,j}+a_{2,j}+\cdots+a_{j,j})\\ &=\sum_{j=1}^n\sum_{i=1}^ja_{i,j}\\ &=\sum_{i=1}^n\sum_{j=1}^ia_{j,i} \end{align*}

edited Jun 22 '21 at 18:44

answered Jun 22 '21 at 17:13

CY Aries

23,393

Ok, I got it. It is just what the comment suggested. Thanks. – Laxmi Narayan Bhandari Jun 22 '21 at 17:36
Please amend your question to include the next step. – CY Aries Jun 22 '21 at 18:04
I have edited the question. Please add the required steps in your answer. – Laxmi Narayan Bhandari Jun 22 '21 at 18:12
Not exactly. But there are some mistakes. Amended now. – CY Aries Jun 22 '21 at 18:33
Shouldn't it be $\sum_{i=1}^j $ instead of $ \sum_{i=1}^i$? – Laxmi Narayan Bhandari Jun 22 '21 at 18:44
yes. lot of mistakes – CY Aries Jun 22 '21 at 18:45
The previous way of first summing with index $i$ was also viable, but no problem. – Laxmi Narayan Bhandari Jun 22 '21 at 18:47

Showing $\sum_{i=1}^n\sum_{j=1}^n\frac1{ij}=\sum_{i=1}^n(\sum_{j=1}^i+\sum_{j=i}^n)\frac{1}{ij}-\sum_{i=1}^n\frac1{i^2}$

1 Answers1