How do I express $a+ab+a^2b+a^2b^2+a^3b^2+a^3b^3+a^4b^3+....$?

Question

How do I express the following infinite sum?

$$a+ab+a^2b+a^2b^2+a^3b^2+a^3b^3+a^4b^3+....$$

I can't quite figure this out. I'm only looking for hints, or maybe the full expression, nothing fancy required!

Answer 1

$$a+\sum_{n\ge 1}(ab)^n(1+a)$$ is one of the possible formulæ. Another would be $$\sum_{n\ge 0}a^{n+1}b^n(1+b).$$

Answer 2

$$a+ab+a^2b+a^2b^2+a^3b^2+a^3b^3+a^4b^3+.... =\sum _{k=1}^\infty a^k(b^{k-1}+b^k)$$

Answer 3

$$\sum_{n=1}^{\infty}a^nb^{n-1} +a^nb^n$$

That's what I first thought when I saw the problem

Answer 4

$1+a+ab+a^2b+...=

$\sum_{k=0}^{\infty}(a^{k+1}b^k+a^kb^k) = (a+1)\sum_{k=0}^{\infty}(ab)^k$.

This sum only converges if $|ab| < 1$ in which case it converges to $\frac {a+1}{1- ab}$.

But your sum is $1$ less. $\sum_{k=0}^{\infty}(a^{k+1}b^k+a^kb^k) = (a+1)(\sum_{k=0}^{\infty}(ab)^k)-1$.

Which converges to $\frac {a+ab}{1-ab} $.