Calculate $\sum_{n\geq 0}\frac{H_n}{10^n}$

Question

Question is like in the title and my attempt is

Let have sequence $$a_n = <\frac{H_0}{10^0},\frac{H_1}{10^1},\frac{H_2}{10^2},\dots>$$ where $H_n$ is n-th harmonic number.

And we have to construct $$b_n = \langle a_0,a_0+a_1,a_0+a_1+a_2,\dots\rangle$$ and it will be our answer.

Let's have generating function for sequence $a_n$: $$A(x) = \sum_{n=0}^\infty a_n x^n$$

And generating function for sequence $b_n$ will be $$B(x) = \frac{A(x)}{1-x}$$

I started making $\frac{H_n}{10^n}$ a bit easier(there are my doubts).

we can rewrite $H_n$ as $$1+\dfrac12+\dfrac13+\dots = 1+\frac{\frac{n(n+1)}{2}-\dfrac22}{n!} = 1+\frac{n(n+1)-2}{2*n!}$$ so now we can rewrite $a_n$ as $$a_n =\langle\frac{1}{10^n}+\frac{n(n+1)-2}{2*n!*10^n}\rangle$$

Generating function for $$\sum_{n=0}^\infty \frac{1}{10^n}x^n = \frac{1}{1-\frac{1}{10}x}$$

But i'm having problem with second part of it... And i'm not sure if i don't overcomplicated this task by making it "easier". I'd like to use generating functions cause it's task for them, but i'd be happy to see some other solutions. Thanks in advance.

Answer 1

You can use the well known identity

$$\sum _{n=0}^{\infty } H_{n}{z}^{n}= \frac{\ln(1-z)}{z-1}. $$

Answer 2

It is known that

$$\sum_{k=1}^{\infty} H_n x^n = -\frac{\log{(1-x)}}{1-x}$$

Plug in $x=1/10$.

I don't think $H_0$ is defined.

EDIT

You can define $H_0=0$ using the expression

$$H_n = \int_0^1 dx \frac{1-x^n}{1-x}$$

Answer 3

It's straightforward to prove the identity by formal manipulation actually. You want $\displaystyle\sum_{n \ge 0} H_n \frac{1}{10^n}$, so let's consider the generating function $A(z) = \sum_{n \ge 0} H_n z^n$, and evaluating $A(\frac1{10})$ will give the answer you want.

By definition, we have $$A(z) = \sum_{n \ge 0} H_n z^n = \sum_{n \ge 0} \sum_{k=1}^{n} \frac{1}{k} z^n = \sum_{k \ge 1} \frac1k \sum_{n \ge k} z^n = \sum_{k \ge 1} \frac1k \frac{z^k}{1-z} = \frac{1}{1-z} \sum_{k \ge 1} \frac{z^k}{k} = \frac{1}{1-z} \ln \frac{1}{1-z}.$$

The last step, the well-known series for $\displaystyle \ln \frac{1}{1-z}$, can itself be derived as follows (interpret $\int (\cdot)$ as $\int_0^z (\cdot)\, dz$):

$$\sum_{k \ge 1} \frac{z^k}{k} = \sum_{k \ge 0} \int z^k = \int \sum_{k \ge 0} z^k = \int \frac{1}{1-z} = -\ln(1-z).$$