What is the generating function for the sequence $\{kH_k\}_{k \geq 1}$ I am trying to answer exercise 3.1 from Sedgewick and Flajolet's "Introduction to the Analysis of Algorithms".  That is, I am trying to find the Ordinary Generating Function for the sequence:

$$\{kH_k\}_{k \geq 1}$$

I think that the generating function for this sequence is:

$$H_1 + 2H_2z + 3H_3z^2 + 4H_4z^3 + ...$$

which abbreviates to

$$\sum_{n \geq 0} (n+1)H_{n+1}z^n $$

By inspection, I see that this has the form of the Index Multiply operation on OGFs.  This means that our generating function can be obtained by differentiating the OGF for the sequence

$$\sum_{n \geq 0}H_n z^n$$

I know that this

$$\sum_{n \geq 1}H_n z^n$$

has a well known OGF. But that is for the range $n \geq 1$, and our sequence has range $n \geq 0$.  If I split off the first term $n = 0$, we get the sum:

$$H_0 + \sum_{n \geq 1}H_n z^n$$

And it's now that I've lost my confidence.  Is the method I've outlined above the right way to go in finding the desired OGF?  If it is OK, it seems to rest on the existence of a zeroth Harmonic number.  Is there actually a zeroth Harmonic number $H_0$?

Thanks.
We know that a generating function for the Harmonic series is: 

$$
\sum_{i =1}^{\infty} H_i z^i = -\frac{\log(1-z)}{1-z}
$$

you may verify it by taking the MacLaurin expansion.

Taking the derivative with respect to $z$:

$$
\frac{d}{dz} \left ( \sum_{i =1}^{\infty} H_i z^i \right ) = \frac{d}{dz} \left ( -\frac{\log(1-z)}{1-z} \right )
$$

yields:
$$
\sum_{i = 1}^{\infty} iH_i z^{i-1} = \frac{1 -\log(1-z)}{(1-z)^2}
$$

Notice that LHS is exactly the formal series you mentioned.