Harmonic numbers identity

[alyafey22]

Prove the following

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

where we define

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

 

[Euge]

Spoiler

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

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

 

[alyafey22]

Spoiler

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.

 

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

 

[CellCoree]

I am always excited to see mathematical identities and proofs. The harmonic numbers identity is a fascinating one that relates the harmonic numbers, H_k, to the generalized harmonic numbers, H^{(k)}_n, and the squared harmonic numbers, H^2_n.

To prove this identity, we will use mathematical induction. First, let's look at the base case n=1:

\sum_{k=1}^1 \frac{H_k}{k} = \frac{H_1}{1} = 1 = \frac{H_1^2 + H_1^{(2)}}{2}

which is true.

Now, assume that the identity holds for some value n. We will show that it also holds for n+1:

\sum_{k=1}^{n+1} \frac{H_k}{k} = \frac{H_{n+1}^2 + H_{n+1}^{(2)}}{2}

Using the definition of H^{(k)}_n and H^2_n, we can rewrite the right side as:

\frac{(H_n + \frac{1}{n+1})^2 + \sum_{j=1}^n \frac{1}{(j+1)^2}}{2}

Expanding the squared term and simplifying, we get:

\frac{H_n^2 + H^{(2)}_n + 2H_n\frac{1}{n+1} + \frac{1}{(n+1)^2} + \sum_{j=1}^n \frac{1}{j^2 + 2j + 1}}{2}

Using the fact that \frac{1}{(n+1)^2} + \sum_{j=1}^n \frac{1}{j^2 + 2j + 1} = \frac{1}{n+1} (see proof below), we can simplify further to get:

\frac{H_n^2 + H^{(2)}_n + 2H_n\frac{1}{n+1} + \frac{1}{n+1}}{2}

which can be rewritten as:

\frac{(H_n + \frac{1}{n+1})^2 + H_n^{(2)}}{2}

And this is equal