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I am trying to find a closed form expression for the series: $$ F(s) = \sum_{n>0} \frac{s^n}{n} \quad \text{for} \quad |s|<1$$

We know that we can define the harmonic number $ h_n = \sum_{i=0}^n f_i$ where $f_0 = 0, f_n = 1/n$ so i tried to use that;

$$ h_n = h_{n-1} + f_n $$

If we multiply both sides of the equation with $s^n$ and sum them both for n>1 we get:

$$\sum_{n>1} h_n s^n = s \sum_{n>1} h_{n-1}s^{n-1} + \sum_{n>1} s^n f_n = s \sum_{n>0} h_{n}s^{n} + \sum_{n>0} \frac{s^n}{n}$$

If we define: $ G(s) = \sum_{n>0} h_n s^n $ we can rewrite this as:

$$G(s) (1-s) = F(s) \rightarrow G(s) = \frac{F(s)}{1-s} $$.

This is where i get stuck, I don't know how I should keep going to find a closed form. If anyone have tips, of can find mistakes in any of my steps any help or guidance would be appreciated.

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Hint, consider integrating a certain function that you know has a power series looking like the derivative of what you want. Think about the function $$1/(1-x)$$

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If you're familiar with power series there are a couple of ways to arrive at the series in question. The most direct is by using the expansion

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} \cdots \qquad \left| x \right| \lt 1 $$

Then a little manipulation: first replacing $\,x\,$ with $\,-x\,$, and using the fact $\,\left| -x \right| = \left| x \right|,\,$ gives us

$$\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4} \cdots \qquad \left| x \right| \lt 1 $$

and then multiplying by $\,-1\,$ yields

$$-\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4} \cdots \qquad \left| x \right| \lt 1 $$

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