Calculate $\sum_{n\geq 0}\frac{H_n}{10^n}$ 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.
 A: 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
$$H_n = \int_0^1 dx \frac{1-x^n}{1-x}$$
A: You can use the well known identity

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

A: 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 $\int_0^z (\cdot)\, dz$):
$$\sum_{k \ge 1} \frac{z^k}{k} = \sum_{k \ge 0} \int z^k = \int \sum_{k \ge 0} z^k = \int \frac{1}{1-z} = -\ln(1-z).$$
