# Trying to sum a finite series

I've tried to solve a financial mathematical task and I need to sum such a finite series:

$$\sum_{k=1}^{6}\frac{1.05^{3k+1}}{1.05^{3k+1}-1}$$

So I decided that maybe I'll try to find a solution to : $$\sum_{j=1}^{n}\frac{x^{j}}{x^{j}-1}$$ My first idea was to $+1$ and $-1$ in the numerator. So I would get: $$\sum_{j=1}^{n}1+\sum_{j=1}^{n}\frac{1}{x^{j}-1}$$ But either it is not the right approach or I just don't see how to continue it. Could somebody help me with the problem? I know I could just put the first formula to mathematica, but I couldn't do such a thing during an exam or a test, so I would like to find a paper answer.

EDIT:

I also need to find a sum of: $$\sum_{j=1}^{n}\frac{jx^{j}}{x^{j}-1}$$ But that's even harder.

• For your first sum you can use mathemateca (or walframalpha online), for your sum it's equal to :$$\sum_{j=0}^{n}\frac{1}{1-x^j}=\sum_{i=0}^{+\infty} k(i)x^i$$ where $k(i)$ is the number of divisors of $i$ Mar 13, 2015 at 23:28
• How do you know? Could you tell me where I can find a proof of it? :) Mar 13, 2015 at 23:36
• @Elaqqad The number of divisors that are at most $n$, actually.
– Did
Mar 13, 2015 at 23:41
• @Novsar Why not try to prove it yourself, now that you have the result?
– Did
Mar 13, 2015 at 23:41
• @Novasar I'm sorry, $k(i)$ the number of divisors of $i$ less than $n$, (another mistake in my comment: start the sum from $j=1$ not $j=0$ ) Mar 13, 2015 at 23:43

$$\sum_{n=1}^{6}\frac{1.05^{3n+1}}{1.05^{3n+1}-1}=\sum_{n=1}^{6}\frac{1}{1-(\frac{20}{21})^{3n+1}}$$ and then use $$\frac{1}{1-x^m}=\sum_{n=0}^{\infty }x^{nm}$$