MHB Harmonic Numbers Identity Proof?

[alyafey22]

Prove the following

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

where we define

$$H^{(k)}_n = \sum_{j=1}^n \frac{1}{j^k} \,\,\, ; \,\,\, H^2_n = \left( \sum_{j=1}^n \frac{1}{j}\right)^2 $$​

 

[Euge]

Spoiler

We have

$$\sum_{k = 1}^n \frac{H_k}{k} = \sum_{1 \le j \le k \le n} \frac{1}{kj}.$$

By symmetry,

$$\sum_{1 \le j \le k \le n} \frac{1}{kj} = \sum_{1 \le k \le j \le n} \frac{1}{kj}.$$

Thus

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

Therefore

$$\sum_{k = 1}^n \frac{H_k}{k} = \frac{H_n^2 + H_n^{(2)}}{2}.$$

 

[alyafey22]

Spoiler

We have

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

You are hiding the crucial steps in the solution.

 

[Euge]

I will make an edit and put more detail in the solution.