Convergence of $S=a(a+b)+a^2(a^2+b^2)+\dots+a^n(a^n+b^n)$ 
Check the convergence of $S_n=a(a+b)+a^2(a^2+b^2)+\dots+a^n(a^n+b^n)$ where $|a|,|b|<1$

I am new to convergence. I tried to find $\lim_{n\to \infty} S_n$. If it is a finite number then the series converges otherwise it diverges. Am I correct? Also how do I find the limit. I tried to simplify and removed the parentheses. The I used the formula for a Geometric Series to help me evaluate the limit. But I cannot understand how to evaluate the limit after this.
 A: If it converges then $a^{2n}(a^{2n}+b^{2n}) \to 0$ which is possible only when $|a| <1$ and $|ab| <1$.  So these inequalities are necessary for convergence. Conversely, if these inequalities hold then it is clear that $S_n$ converges by comparison with the geometric series $\sum |a|^{2n}$ and $\sum |ab|^{n}$. 
Hence a necessary and sufficent condition for existence of $\lim S_n$ is $|a| <1$ and $|ab| <1$. 
Also $\lim S_n=\frac {a^{2}} {1-a^{2}}+\frac {ab} {1-ab}$.
A: Let $A=a^2$ and $B=ab$.
$$S_n=a(a+b)+a^2(a^2+b^2)+\dots+a^n(a^n+b^n)$$
can be written :
$$S_n=A+A^2+...+A^n+B+B^2+...+B^n$$
i.e.,
$$S_n=A\dfrac{1-A^n}{1-A}+B\dfrac{1-B^n}{1-B}$$
As $|A|<1$ and $|B|<1$, when $n \to \infty$, $S_n$ converges to :
$$S=A\dfrac{1}{1-A}+B\dfrac{1}{1-B}=a^2\dfrac{1}{1-a^2}+ab\dfrac{1}{1-ab}$$
A: You are right about the definition : a serie converges iff the limit of partial sums converges.
but for finding this limit :
1- when you remove parantheses, you have two kind of algebraic formulas : one in terms of "a" and the other one in terms of "a" and "b" (and both are geometric series!)
2- separately, evaluate the limit of these two geometric serie(of course, both are going to be dependant to "a" and "b" respectively!)
3- limit of sum of two sequence is equal to sum of their limits. adding the results of last part will determine convergence of your serie.
A: Applying the standard rule for the sum of a geometric progression you get:
$$\begin{equation} \begin{aligned}
S_n
&= a(a+b)+a^2(a^2+b^2)+\dots+a^n(a^n+b^n) \\[6pt]
&= \sum_{i=1}^n a^i (a^i + b^i) \\[6pt]
&= \sum_{i=1}^n (a^2)^i + \sum_{i=1}^n (ab)^i \\[6pt]
&= a^2 \cdot \frac{1 - (a^2)^{n}}{1-a^2} + ab \cdot \frac{1 - (ab)^{n}}{1-ab} \\[6pt]
\end{aligned} \end{equation}$$
Hence, taking $|a|<1$ and $|b|<1$ you have $(a^2)^{n} \rightarrow 0$ and $(ab)^{n} \rightarrow 0$, which gives the limit:
$$S_\infty \equiv \lim_{n \rightarrow \infty} S_n
= \frac{a^2}{1-a^2} + \frac{ab}{1-ab}.$$
