Prove that that $S_n = 1+\lambda+\lambda^2+...+\lambda^n$ converges if $0 \le \lambda \lt 1$, $\lambda\in\mathbb{R}$ 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 $\delta &gt; 0$, then $\lambda^{n}/(1-\lambda) &lt; \delta$ iff $\lambda^{n} &lt; \delta (1-\lambda)$, iff $n|\log \lambda| &gt; |\log \delta(1-\lambda)|$. (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 $0&lt;q&lt;1$ then we can easily verify that $S_n \to (1-q)^{-1}$ as $n\to \infty$.  

Fix $\varepsilon &gt; 0$, 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 $|q^n|\to 0$?