Sum of sum of factors of numbers that are < 1

Question

Let $a, b \in (0,1)$. Given an integer $N$, how can I calculate the following sum: $$ \sum_{i=1}^N{\sum_{j=0}^i}a^j\cdot b^{i-j} $$

I tried different tricks so far, but none seemed to solve it.

If it simplifies things, it is possible to consider only the limit: $$ \lim_{N\rightarrow \infty} \sum_{i=1}^N{\sum_{j=0}^i}a^j\cdot b^{i-j} $$

Thanks!

Answer 1

Write $x = \frac ab$. We have $$ \begin{align*} \sum_{i=1}^N{\sum_{j=0}^i}a^j\cdot b^{i-j} &= \sum_{i=1}^Nb^i{\sum_{j=0}^i}x^j\\ &= \sum_{i=1}^Nb^i \frac{x^{i+1} - 1}{x - 1}\\ &= \frac1{x-1}\left(x\sum_{i=1}^N b^ix^i - \sum_{i=1}^Nb^i\right), \end{align*} $$ and now it's a matter of evaluating geometric series, one with ratio $bx = a$ and one with ratio $b$.

Answer 2

Find a close formula $f(i)$ for $$\sum_{j=0}^{i}a^j\cdot b^{i-j}=a^0\cdot b^i+a^1\cdot b^{i-1}+a^2\cdot b^{i-2}+...+a^i\cdot b^0$$ Note that this is a geometric series.
Then try to do the same with $$\sum_{j=1}^Nf(i)$$