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Is the left hand side equal to the right hand side here? If so, why?

$$\sum_{i=1}^n \frac{1}{i} + 2\cdot\sum_{1\le i\lt j\le n}\frac{1}{ij} = \left(\sum_{i=1}^n \frac{1}{i}\right)^{\!2}+\sum_{i=1}^n \frac{1}{i} - \sum_{i=1}^n \frac{1}{i^2}$$

Thanks!

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  • $\begingroup$ The $\sum_{i=1}^n 1/i$ cancels out on both sides. So your problem is equivalent to asking if $$2\cdot\sum_{1\le i\lt j\le n}\frac{1}{ij} = \left(\sum_{i=1}^n \frac{1}{i}\right)^{\!2}- \sum_{i=1}^n \frac{1}{i^2}. $$ $\endgroup$ Nov 26 '19 at 16:30
  • $\begingroup$ Note that the second term on the LHS is half the sum of all fractions 1/ij where 1<= i,j <= n, removing the diagonals. In other words, the removal of when i=j $\endgroup$
    – Gabe
    Nov 26 '19 at 16:31
  • $\begingroup$ why would it be half? $\endgroup$
    – Jdo
    Nov 26 '19 at 16:36
  • $\begingroup$ It's half because j>i. This eliminates the other half of fractions where i>j $\endgroup$
    – Gabe
    Nov 26 '19 at 16:40
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Given: $$\sum_{i=1}^n \frac{1}{i} + 2\cdot\sum_{1\le i\lt j\le n}\frac{1}{ij} = \left(\sum_{i=1}^n \frac{1}{i}\right)^{\!2}+\sum_{i=1}^n \frac{1}{i} - \sum_{i=1}^n \frac{1}{i^2}$$ Obviously you can write it as $$2\cdot\sum_{1\le i\lt j\le n}\frac{1}{ij} = \left(\sum_{i=1}^n \frac{1}{i}\right)^{\!2} - \sum_{i=1}^n \frac{1}{i^2}$$ Now, notice that $$2\cdot\sum_{1\le i\lt j\le n}\frac{1}{ij} + \sum_{i=1}^n \frac{1}{i^2} = \sum_{1\le i, j\le n}\frac{1}{ij}$$ This can be visualized with a picture, excuse my bad drawing skills. Graph of when i,j <= 5

Notice how the sum of the red and green areas is equal to the first term of the LHS and the blue area is equal to the second term of the LHS. The total graph is the sum of all fractions $\frac{1}{ij}$

Now, all that’s left is to prove $$\sum_{1\le i, j\le n}\frac{1}{ij}=\left(\sum_{i=1}^n \frac{1}{i}\right)^{\!2}$$

The RHS can be expanded into the LHS easily just by doing out the square (try it), so the proof is done.

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The answer is yes, and here's why: $$\left(\sum_{i=1}^n\frac1i\right)^{\!2}=\left(\sum_{i=1}^n\frac1i\right)\left(\sum_{j=1}^n\frac1j\right)=\sum_{i=1}^n\sum_{j=1}^n\frac1{ij}. $$ This is analogous to a matrix entry $A_{ij},$ with $A$ symmetric. So, if you subtract out the main diagonal, which you can write as $$\sum_{i=1}^n\frac{1}{i^2}, $$ you're left with the off-diagonal elements. But, because $A_{ij}=A_{ji},$ the difference $$\sum_{i=1}^n\sum_{j=1}^n\frac1{ij}-\sum_{i=1}^n\frac{1}{i^2} $$ can be re-written using the notation $$2\cdot\sum_{1\le i\lt j\le n}\frac{1}{ij}, $$ as required, which is also equivalent to $$\sum_{\begin{array}{c}i,j=1\\i\not=j\end{array}}^n\frac{1}{ij}. $$

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Second sum on LHS (including the 2) is equivalent to $$\sum_{i, j} \frac{1-\delta_{ij}}{ij}$$ so you get the two sums on the RHS: $$\sum_i\sum_j\frac1{ij}= \sum_i\sum_j\frac1i\frac1j= \sum_i\frac1i\sum_j\frac1j=\left(\sum_{i}\frac1i\right)^2$$

$$\sum_i\sum_j\frac{\delta_{ij}}{ij}= \sum_i\frac1{i^2}$$

The two goes away because we already double-count with the two sums written as I did, considering both the cases $i>j$ and the $j>i$ which are equivalent as the term $\frac1{ij}$ is symmetric under the exchange $i\leftrightarrow j$.

As someone already pointed out, you have the same sum on both sides, so it cancels out;

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