Harmonic Numbers Identity Proof?

  • #1
alyafey22
Gold Member
MHB
1,561
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Prove the following

\(\displaystyle \sum_{k=1}^n \frac{H_k}{k} = \frac{H_n^2+H^{(2)}_n}{2}\)​

where we define

\(\displaystyle H^{(k)}_n = \sum_{j=1}^n \frac{1}{j^k} \,\,\, ; \,\,\, H^2_n = \left( \sum_{j=1}^n \frac{1}{j}\right)^2 \)​
 
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  • #2
We have

\(\displaystyle \sum_{k = 1}^n \frac{H_k}{k} = \sum_{1 \le j \le k \le n} \frac{1}{kj}.\)

By symmetry,

\(\displaystyle \sum_{1 \le j \le k \le n} \frac{1}{kj} = \sum_{1 \le k \le j \le n} \frac{1}{kj}.\)

Thus

\(\displaystyle 2 \sum_{1 \le j \le k \le n} \frac{1}{kj} = \sum_{1 \le j,\, k \le n} \frac{1}{kj} + \sum_{1 \le j,\,k \le n, k = j} \frac{1}{kj} = \left(\sum_{k = 1}^n \frac{1}{k}\right)^2 + \sum_{k = 1}^n \frac{1}{k^2} = H_n^2 + H_n^{(2)}.\)

Therefore

\(\displaystyle \sum_{k = 1}^n \frac{H_k}{k} = \frac{H_n^2 + H_n^{(2)}}{2}.\)
 
Last edited:
  • #3
Euge said:
We have

\(\displaystyle \sum_{k = 1}^n \frac{H_k}{k} = \sum_{1 \le j \le k \le n} \frac{1}{kj} = \dfrac{\left(\sum_{k = 1}^n \frac{1}{k}\right)^2 + \sum_{k = 1}^n \frac{1}{k^2}}{2} = \frac{H_n^2 + H_n^{(2)}}{2}.\)

You are hiding the crucial steps in the solution.
 
  • #4
I will make an edit and put more detail in the solution.
 

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