# Sum of the recripocals of the Harmonic Numbers

Let $H_n=1+\frac12 +\frac13 +\cdots + \frac1n$. I want to show that the infinite sum $$\sum_{n=1}^{\infty} \dfrac{1}{H_n}$$ diverges.

My approach:

For $n,m\in\mathbb{N}$, is easy to verify that $H_n<H_m$ if $n<m$, thus $H_n<H_{2^n}$ for every $n\in\mathbb{N}$. But, \begin{align*} H_{2^n}&=1+\frac12 +\frac13 +\cdots + \dfrac{1}{2^n}\\ &<1+\left(\frac12+\frac12\right)+\left(\frac14+\frac14+\frac14+\frac14 \right)+\cdots+ \underbrace{\left(\dfrac{1}{2^{n-1}}+\cdots+\dfrac{1}{2^{n-1}}\right)}_{2^{n-1}\text{ times}}\\ &=1+1+\cdots +1=n \end{align*}

Then, $H_{2^n}\leqslant n$ for every $n$, so $$\frac1n\leqslant \dfrac{1}{H_n},$$ for every $n$ and since $$\sum_{n=1}^{\infty}\frac1n$$ diverges, the infinite sum $$\sum_{n=1}^{\infty}\dfrac{1}{H_{n}}$$ diverges too.

Am I right?

• You can show directly that $H_n = 1 +\frac 12 + \cdots + \frac 1n < 1 + 1+ \cdots + 1 = n$. (And your proof is correct) – user99914 Oct 17 '15 at 22:46
• And I just realise of that... Thanks. – Juan C. Cala Oct 17 '15 at 22:55

$\displaystyle\frac 1 {H_n} > \frac 1 n$, so divergence follows from the basic comparison test.

• And you say that without using no arguments?... If you read, that's actually what I've done in my approach. – Juan C. Cala Oct 17 '15 at 22:40
• @Jccalab : I didn't intend to present a complete answer, just something pointing you in the right direction, especially in view of the fact that there's not much to do after that. Here's an argument: $\displaystyle 1+ \frac 1 2+\frac 1 3+\frac 14+\cdots+\frac 1 n \vphantom{\frac\int{\displaystyle\int}}$ is a sum of $n$ terms. All of them except the first are less than $1$. Therefore the sum is less than $\displaystyle\underbrace{1 + 1 + 1 + \cdots + 1}_{n\text{ terms}} = n$. Since $H_n$ is thus shown to be less than $n$, we get $\displaystyle \frac 1 {H_n} > \frac 1 n$. ${}\qquad{}$ – Michael Hardy Oct 18 '15 at 3:26
• Your argument is more complicated than it needs to be. You 've show an inequality only for the case when the number of terms is a power of $2$, and you've included more steps than are needed to get to the conclusion that the terms in the sum $H_n$ are small. ${}\qquad{}$ – Michael Hardy Oct 18 '15 at 3:30