Summing $\sum_{n=1} ^{\infty} \frac{n}{n+ 1}(x^{n + 1})$

Question

Assuming $x < 1$, what is the sum of, $$\sum_{n=1} ^{\infty} \frac{n}{n+ 1}(x^{n + 1}) = \frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 + \frac{4}{5}x^5 + \dots $$

My attempt so far:

$$ln(1 - x) = -x - \frac{(-x)^2}{2} + \frac{(-x)^3}{3} - \frac{(-x)^4}{4} \dots $$

$$= -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} \dots $$

$$\Longrightarrow -ln(1 - x) - x = \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4}\dots $$

Here things begin to grind to a halt.

If I were to multiply the right hand side by $(1 + x + x^2 + \dots)$ I'd have,

$$(1 + x + x^2 + \dots)(\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots) = \frac{x^2}{2} + (\frac{x^3}{2} + \frac{x^3}{3}) + (\frac{x^4}{2} + \frac{x^4}{3} + \frac{x^4}{4}) + \dots $$

This is obviously wrong, but it feels like the sort of thing to do. If only the fractions of the RHS of the above equation were$ \frac{1}{2} + (\frac{1}{3} + \frac{1}{3}) + (\frac{1}{4} + \frac{1}{4} + \frac{1}{4}) + \dots$ I'd be done.

I've also tried adding or multiplying by $e$ or multiples thereof, but summing fractions of factorials doesn't seem to be fruitful. Have tried adding other log series to this sum, but so far no amount of grouping gives me one additional fraction for each term.

Any hints? Solving this without resorting to any techniques from calculus is part of the challenge.

Answer 1

Observe that $$ \sum_{n = 1}^\infty \frac{n}{n + 1}x^{n + 1} = \sum_{n = 1}^\infty x^{n + 1} - \sum_{n = 1}^\infty \frac{x^{n + 1}}{n + 1}. $$ The first sum looks like a geometric series and the second sum looks like the power series of a logarithmic function. Can you take it from here?

Answer 2

Let's just play around with the coefficients. You say you want no use of differentiation or integration, but I find that strange since you have to arrive at the series for $\ln$ somehow...

$$\sum_{n\ge1}\frac{n\color{red}{+1-1}}{n+1}x^{n+1}=\sum_{n\ge1}x^{n+1}-\sum_{n\ge1}\frac{1}{n+1}x^{n+1}=\sum_{n\ge2}x^n-\sum_{n\ge1}(-1)^n\frac{(-x)^n}{n}+x$$

Can you continue?

Answer 3

Another approach:

$$\frac n{n+1}x^{n+1}=\int n x^n\,dx=x\int\left(x^n\right)dx$$

and with this, for $\;|x|<1\;$ :

$$\sum_{n=1}^\infty\frac n{n+1}x^{n+1}=\int\left(\sum_{n=1}^\infty nx^n\right)dx=\int\left(x\left(\sum_{n=0}^\infty x^n\right)'\right) dx$$

Observe that the middle term can be taken as sum from $\,n=0\;$ ...

and now:

$$\sum_{n=0}^\infty x^n=\frac1{1-x}\implies\left(\sum_{n=0}^\infty x^n\right)'=\frac1{(1-x)^2}$$

and thus ( again, for $\;|x|<1\;$):

$$\int\left(x\left(\sum_{n=0}^\infty x^n\right)'\right) dx=\int\frac x{(1-x)^2} dx=\int\left(-\frac1{1-x}+\frac1{(1-x)^2}\right) dx=\log(1-x)+\frac1{1-x}+C$$