Let $$S = \sum\limits_0^\infty x^j,$$ where $|x|<1$. I want to show by induction that $$S=\frac{1}{1-x}.$$ But I'm stuck in my attempts to come up with an induction definition.
For example, $S_j = \frac{1}{1-x}-x^j$ clearly does not work here. Should I seek a recursive definition instead?
You should show that for every finite $N$ from induction that \begin{align} \sum_{k=0}^{N} r^k = \frac{1-r^{N+1}}{1-r} \end{align} then it follows that this limit converges if and only if $r^{N+1}$ converges to a finite non-1(so that the denominator is well defined) value, which happens if and only if $|r| < 1$.
But here's a much more short proof: \begin{align}S_{N}=\sum_{k=0}^{N} r^k = 1 + r + r^2 + ... + r^N \end{align} \begin{align} (1-r)S_N = 1 + r + r^2 +... + r^N - r - r^2 - ... -r^{N+1} = 1 - r^{N+1} \end{align} This implies that $S_N = \frac{1-r^{N+1}}{1-r}$. Then let $N \rightarrow \infty$ to see it converges if and only if $|r| < 1$