My friend asked me a question about this (from Russia IMO or something I forgot). The question asked for \(n=2013\). I know the answer by just trying small numbers and then predict. Here's the question to prove:

Prove that \(\sum\limits_{j=1}^n (\sum\limits_{i=j}^n \frac{1}{i})^{2} + \sum\limits_{i=1}^n \frac{1}{i} = 2n\)

If you're confused with these stuffs, it is...

\((\frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n})^{2} + (\frac{1}{2} + ... + \frac{1}{n})^{2} + ....... + (\frac{1}{n})^{2} + \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} = 2n\)

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TopNewestLet \( H_n = \sum_{i=1}^n \frac1{i} \) (with the convention that \( H_0 = 0 \) ). My approach was to write the quantity on the left side as \( S_n = \sum_{j=1}^n (H_n - H_{j-1})^2 + H_n \). It is enough to show that \( S_{n+1}-S_n = 2 \), since \( S_1 = 2 \) is the base case for an easy induction.

This difference equals \( \sum_{j=1}^{n+1} \left( (H_{n+1}-H_{j-1})^2-(H_n-H_{j-1})^2 \right) + \frac1{n+1} \), which simplifies by a difference of squares formula to \( \sum_{j=1}^{n+1} \frac1{n+1} ( H_{n+1} + H_n - 2 H_{j-1}) + \frac1{n+1} \), which becomes \( H_{n+1} + H_n + \frac1{n+1} - \frac2{n+1} \sum_{j=1}^{n+1} H_{j-1} = 2 H_{n+1} - \frac2{n+1} \sum_{j=1}^{n+1} H_{j-1} \).

Ah, but now we can use the identity \( \sum_{i=0}^n H_i = (n+1)H_n - n \), which is easy to prove by induction. (I found it on the wikipedia page for "harmonic number.") Plugging this in, we get \( 2 H_{n+1} - \frac2{n+1} ((n+1) H_n - n) = 2 H_{n+1} - 2 H_n + \frac{2n}{n+1} = \frac2{n+1} + \frac{2n}{n+1} = 2 \) as desired.

There is probably a much more elegant way to do the problem (i.e. directly without induction), but this certainly works.

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I don't quite understand how it is enough to prove that \(S_{n+1} - S_{n} = 2\). I understand the rest but I'm just stuck at the first part. =_=" Sorry about that *cries

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\(S_{n+1} - S_{n} = 2(n+1) - 2n = 2\)

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Thank you for the proof. ^__^

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This is from IMC 2013. :)

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Thanks! ^__^

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