I need to prove this using the $\epsilon - N$ definition of convergence.
I know from the definition that $\forall \epsilon \gt 0, \exists N \in\mathbb{R}$ such that $\forall n \gt N, |S_n - L | \lt \epsilon$, where $L = \frac{1}{1-\lambda}$ in this case.
How do I find such an $N$?
We have $S_{n} = (1-\lambda^{n})/(1-\lambda)$ for all $n$; $|S_{n}-L| = | \frac{{-\lambda}^{n}}{1-\lambda}| \leq \frac{\lambda^{n}}{1-\lambda}$ for all $n$. If $\delta > 0$, then $\lambda^{n}/(1-\lambda) < \delta$ iff $\lambda^{n} < \delta (1-\lambda)$, iff $n|\log \lambda| > |\log \delta(1-\lambda)|$. (Note that $\lambda$ and $1-\lambda$ are such that $\log$ is negative.) Can you do the rest from here on?
From the following Formula for finite power series, we can easily see that $$S_n=\sum_{q=0}^n \frac{1-q^{n+1}}{1-q}$$
now if $0<q<1$ then we can easily verify that $S_n \to (1-q)^{-1}$ as $n\to \infty$.
Fix $\varepsilon > 0$, then
$$\left |S_n-L \right|= \left| \frac{1-q^{n+1}-1}{1-q} \right | = \left|\frac{1}{1-q}\right| \left |q^{n+1} \right|=c\left |q^{n+1} \right|$$
where $c$ is the constant term $(1-q)^{-1}$.
Do you think you can show that $|q^n|\to 0$?
