I want to show that $$\sum_{k=1}^{n} (5x)^k$$ converges uniformly. That is, it converges uniformly to $\sum_{k=1}^{\infty} (5x)^k$ for $|x| < a <\frac{1}{5}$. Following the idea of the Weierstrass M-test, consider all $x$ such that $$|x| < a < \frac{1}{5} \Rightarrow |5x| < 5a < 1 $$, so we have
$$ \begin{align} | \sum_{k=1}^{n} (5x)^k - \sum_{k=1}^{\infty} (5x)^k| & = |\sum_{k=n+1}^{\infty} (5x)^k| \\ &< \sum_{k=n+1}^{\infty} |(5x)^k| \\ &< \sum_{k=n+1}^{\infty} (5a)^k \\ &< \epsilon \end{align} $$
for any given $\epsilon$, since $\sum_{k=n+1}^{\infty} (5a)^k$ converges for $5a < 1$. So, since the $N$ we get for which $n \geq N \Rightarrow | \sum_{k=1}^{n} (5x)^k - \sum_{k=1}^{\infty} (5x)^k| < \epsilon$, is independent of $x$, since $\sum_{k=n+1}^{\infty} (5a)^k$ is independent of $x$, thus our series converges uniformly. So far so good. (Please correct me if I have done something wrong up until now.)
Now, because it's not really clear to me exactly what's happening here I want to put numbers into this proof and say for $\epsilon = 0.01$ find the $N$ value. And then for a couple values of $x$ verify that indeed $| \sum_{k=1}^{n} (5x)^k - \sum_{k=1}^{\infty} (5x)^k| < 0.01$. But it's at that point that I don't know how to proceed.
Any help would be appreciated.
Edit 1: Changed the range of conversion from $|x| < \frac{1}{5}$ to $|x| < a <\frac{1}{5}$
Hint
Note that $$\sum_{k=n+1}^{\infty} (5x)^k={(5x)^{n+1}\over 1-5x}$$when $|x|<{1\over 5}$.