Double Sigma with nested index, $i$

Question

I am trying to solve this double sigma but my answer doesn't seem right. $$\sum\limits_{i=1}^n \sum\limits_{j=i}^{n^2}1=-i\sum\limits_{i=1}^n n^2 = -in^3$$

Note that the symbols $i$ and $j$ have no meaning outside of the summation. — Alex Provost, Apr 05 '16 at 14:09

Ant · Answer 1 · 2016-04-05T14:12:04.697

1

Second term should be $$\sum_{i=1}^n (n^2 - i+1) = n^3+n - \sum_{i=1}^n i = n^3+n - \frac{n(n+1)}2$$

edited Apr 05 '16 at 14:12

answered Apr 05 '16 at 14:09

Ant

21,098

You have an off by one error. Double check that index on the inner sum – Stella Biderman Apr 05 '16 at 14:11
@stellabiderman thanks! Corrected :-) – Ant Apr 05 '16 at 14:12

score 1 · Accepted Answer · answered Apr 05 '16 at 14:10

Yeah, I'm not sure what you're doing, but if you wind up with the index of summation outside the summation you're doing something wrong. "$i,j$" are not variables here, they are placeholder symbols that is used to compactify notation. As a result, they have no meaning on their own and can't exist outside the summation. This is similar to how the "dummy variable" of integration isn't a real variable and can't exist outside the integral.

Isolating the inside give us $$\sum_{j=i}^{n^2}j^0=n^2-i+1$$ and now we plug this into the outside sum to get $$\sum_{i=1}^{n}(n^2-i+1)=n^3-\frac{n^2+n}{2}+n$$

Double Sigma with nested index, $i$

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