Proving that alternative of convergence of Cauchy series is true or false

Question

I'm trying to prove or disprove that statement $$(\forall \varepsilon > 0)(\exists n_0 \in \mathbb{N})(\forall n \in \mathbb{N}, n > n_0)(\forall p \in \mathbb{N})\left(\left|\sum_{k = n + 1}^{n + p} a_k\right| < \varepsilon \right),$$ is equivalent (i.e. $\iff$) with statement $$(\forall \varepsilon > 0)(\exists n_0 \in \mathbb{N})(\forall p \in \mathbb{N})\left(\left|\sum_{k = n_0 + 1}^{n_0 + p} a_k\right| < \varepsilon \right),$$ but I don't know how to do so, because I have no experience in writing original proofs. What is the way to prove it?

Answer 1

Hint: One direction of the proof is easy because $n_0+1$ is greater than $n_0$, so the second statement is a special case of the first statement.

For the other direction, observe the following: Let $n>n_0$ and $p>0$, then $$ \sum_{k=n+1}^{n+p}a_k=\sum_{k=n_0+1}^{n+p}a_k-\sum_{k=n_0+1}^na_k $$ Therefore, $$ \left|\sum_{k=n+1}^{n+p}a_k\right|=\left|\sum_{k=n_0+1}^{n+p}a_k-\sum_{k=n_0+1}^na_k\right|\leq\left|\sum_{k=n_0+1}^{n+p}a_k\right|+\left|\sum_{k=n_0+1}^na_k\right|<2\varepsilon. $$ Thus, if you replace $\varepsilon$ with $\varepsilon/2$ in the second statement, you can get the first statement (without going through the details of the Cauchy proof itself).