Series from $n=k$ to infinity. What is limit as $k$ goes to infinity? Let $b_k=\sum_{n=k}^\infty a_n$. Suppose $\sum_{n=1}^\infty a_n < \infty$ with $a_n \geq 0$ for all $n$. I want to show $\lim_{k \rightarrow \infty} b_k=0.$ It seems obvious, but I'd like to be formal.
By divergence test, we know $\lim_{n \rightarrow \infty} a_n =0,$ so for any $\varepsilon > 0,$ there is an $N$ such that $a_n < \frac{\varepsilon}{2^n}$ for all $n>N.$ Hence,
$$ \lim_{k \rightarrow \infty}b_k = \lim_{k \rightarrow \infty} \sum_{n=k}^\infty a_n \leq \sum_{n=1}^\infty a_n < \sum_{n=N+1}^\infty \frac{\varepsilon}{2^n} < \sum_{n=1}^\infty \frac{\varepsilon}{2^n}=\varepsilon.$$
Since $\varepsilon$ is arbitrary, we are done. Is this acceptable?
 A: No.  Your statement $a_n\lt \frac{\epsilon}{2^n}$ for all $n \gt N$ is not true.  A direct proof is using the fact that the series is convergent.  Let $s_n=\sum_{k=1}^{n-1} a_k$ while $s=\sum_{k=1}^\infty a_k$.  Since the series is convergent for any $\epsilon \gt 0$ there exists an $N$ so that for all $n\gt N$, $|s-s_n|=|b_n| \lt \epsilon$
A: 
there is an $N$ such that $a_n < \frac{\varepsilon}{2^n}$ for all $n>N.$

This isn't true, because the expression you are bounding $a_n$ by can't be allowed to depend on $n$ when using the definition of limit. For instance the sum with $a_n = \frac{1}{n^2}$ converges, but for any $\varepsilon$  you will find that $\frac{\varepsilon}{2^n}$ is eventually much smaller than $\frac{1}{n^2}$.
Instead, use the fact that the $b_k$s are the difference between the infinite sum and the partial sums that converge to that infinite sum.
A: I mean this is really weirdly stated  for one you don't actually state what $ a_n$ is luckily it doesn't actually matter all that much the assuming that $\lim_{n \rightarrow \infty} a_n =0$ is true and $ a_n < \infty $ for all n.
This tells us that $ \sum_{n=1}^{\infty} a_n $ converges other wise stated there is a number $k$ s.t for all $n >k $  the sequence $a_n$ becomes all $0$'s picking $n=k$ in your first statement this shows that $b_k= \sum_{k}^{\infty} a_k$ is just a sequence of all $0$'s so it must converge to $0$. No epsilons, deltas or anything fancy needed just wielding the definitions.
