A series $\sum_{n=1}^{\infty} a_{n}$ converges if the sequence $s_{n} = \sum_{k=1}^{n} a_{k}$ converges. Suppose that $a_{n} \geq 0$ for all $n\geq 1$. Prove the following are all equivalent:
(i) $\sum_{n=1}^{\infty} a_{n}$ converges.
(ii) the sequence $s_{n} = \sum_{k=1}^{n} a_{k}$ is bounded above.
(iii) for all $\epsilon >0$, there is an $N$ such that $\sum_{k=N+1}^{m} a_{k} < \epsilon$ for all $m > N$.
Edit: So here’s what I’ve done so far:
$\space\space\space\space\space\space\space\space\space\space$ Assume statement $(i)$ is true. Let $\epsilon=1>0$, let $s_{n}= \sum_{k=1}^{n} a_{k}$, and let $L = \sum_{n=1}^{\infty} a_{n}$. Then by
the definition of a limit, since the sum converges, $\exists N, n\geq N \Rightarrow |s_{n}-L| < \epsilon = 1$. Thus, we have
$-1 < s_{n} - L < 1$. By the triangle inequality, $|s_{n} -L| \geq |s_{n}| - |L|$. Since
$a_{n} \geq 0 \forall n \geq 1$, $L \geq 0 \Rightarrow |L| = L$. Thus, $|s_{n}| -L \leq |s_{n}-L| < \epsilon=1 \Rightarrow |s_{n}| < L + 1$. Let
$R = 1+ max\{L+1, s_{1},s_{2}, ..., s_{n}\}$. Then $s_{n} < R \space \forall n\geq 1$, so the sequence is bounded above (i.e.
statement $(i) \Rightarrow$ statement $(ii)$).
$\space\space\space\space\space\space\space\space\space\space$ Assume $s_{n}= \sum_{k=1}^{n} a_{k}$ converges. By the limit definition, we have that
$\forall \epsilon >0, \exists N, \forall n \geq N, |s_{n}-L| = |\sum_{k=1}^{n} a_{k} - \sum_{k=1}^{\infty}a_{k}| = |\sum_{k=n+1}^{\infty} a_{k}| = \sum_{k=n+1}^{\infty} a_{k} <\epsilon$ (the last
step uses the fact that $a_{n} \geq 0 \space \forall n\geq 1$). But since $\sum_{k=n+1}^{\infty} a_{k} > \sum_{k= n+1}^{m}a_{k}$, where $m> n \geq N$,
we have that statement $(i) \Rightarrow$ statement $(iii)$.
$\space\space\space\space\space\space\space\space\space\space$ Now assume statement $(iii)$ is true. Since for all $\epsilon >0$, there is an $N$ such that
$\forall m >N \sum_{k=N+1}^{m} a_{k} < \epsilon \space \Rightarrow s_{N} +\sum_{k=N+1}^{m} a_{k} < s_{N} + \epsilon \space$. But this means that
$ \forall \epsilon >0, L = s_{N} + \sum_{k=N+1}^{\infty} a_{k} < s_{N} + \epsilon$. Since $a_{n} \geq 0\forall n\geq1, s_{N} \leq s_{n} \forall n \geq N$. So we have that
$\forall \epsilon >0, \forall n \geq N (L = s_{N} + \sum_{k=N+1}^{\infty} a_{k} < s_{N} + \epsilon \leq s_{n} + \epsilon)$. Rearranging the equation gives
$\forall \epsilon >0, \exists N, \forall n \geq N \space(L - s_{n} = |L -s_{n}| = |s_{n}-L| < \epsilon)$, which satisfies the definition of a
limit and so statement $(iii) \Rightarrow$ statement $(i)$, creating a loop. Thus, the statements are all equivalent.
I’m not sure if I’m missing something. Any help would be very much appreciated!!!
Hints :
i) -> iii) : $s_n$ converges, then it is a Cauchy sequence.
ii) -> i) : what can you say about the variations of $s_n$ ?
iii) -> ii) : Let $\epsilon = 1$, what can you deduce ?