Compute the sum (S):
$\frac11.\frac12 + \frac12.\frac13 + \frac13.\frac14 + \frac14.\frac15 ... infinite terms + $
$\frac11.\frac12 + \frac12.\frac13 + \frac13.\frac14 + \frac14.\frac15 ... infinite terms + $
.......... $ + \frac12.\frac13 + \frac13.\frac14 + \frac14.\frac15 ... infinite terms + $
......................... $ + \frac13.\frac14 + \frac14.\frac15 ... infinite terms + $
........................................ $ + \frac14.\frac15 ... infinite terms + $
I have tried summing it column-wise and row-wise both and I get different answers both ways.
Row-wise:
each row is $\sum_{r=r}^{r=\inf}\frac{1}{(r)(r+1)}$ where $r$ is $1,1,2,3,4,5...$ for row1,row2,row3,row4,row5...
$\sum_{r}^{\inf}\frac{1}{(r)(r+1)} = \sum_{r}^{\inf}(\frac1r - \frac{1}{(r+1)}) = \frac1r$
$S_r = \frac11 + \frac11 + \frac12 + \frac13 + \frac14 + ...$
Column-wise:
columns can be added to reduce the sum (S) to $S = 2.\frac11.\frac12 + 3.\frac12.\frac13 + 4.\frac13.\frac14 + 5.\frac14.\frac15 ... infinite terms $
$S_c = \frac11 + \frac12 + \frac13 + \frac14 + ...$
$S_r$ and $S_c$ differ by 1 which is clearly an error on my part somewhere. Please help me understand what I am doing wrong? My guess is that since both $S_c$ and $S_r$ diverge to $\inf$, maybe I'm doing a senseless thing trying to equate two infinities