Why $\sup\limits_{n\ge 1} \sum_{k=1}^n a_k = \lim\limits_{n\to\infty} \sum_{k=1}^n a_k$?

Question

If $\{a_n\}$ is a positive numbers sequence, then $\sup\limits_{n\ge 1} \sum_{k=1}^n a_k = \lim\limits_{n\to\infty} \sum_{k=1}^n a_k$. Is this wrong or right? And why?

Answer 1

Hint: Remember that every statement about an infinite series is really a statement about its sequence of partial sums. If $a_k>0$ for all $k$, then what can you say about the sequence of partial sums?

Answer 2

Notice that the sequence $s_n=\sum\limits_{i=1}^n a_n$ is increasing.

We prove that if $s_n$ is an increasing sequence then $\sup\limits_{n\geq 1}s_n=\lim\limits_{n\to \infty}s_n$.

Case $1$: $\sup\limits_{n\geq 1}s_n=\infty$, then the set of values taken by $s_n$ is not bounded above, so for any $m\in \mathbb N$ we can find an $N$ with $s_N>m$, since $s_n$ is increasing we have $s_n>m$ for any $n>N$

Case $2$: $\sup\limits_{n\geq 1}s_n=l$, by the property of the supremum, for any $\epsilon>0$ we can find an $N$ so that $S_N\in(l-\epsilon,l]$. Since $s_n$ is increasing and $l$ is an upper bound we have $s_n\in(l-\epsilon,l]$ for $n\geq N$. So the limit is in fact $l$.

Answer 3

Recall the following:

Any bounded increasing sequence sequence converges to is supremum.

Since the $a_k$ are all positive, the sequence $\{\sum_{k=1}^na_k\}$ is increasing. Thus you only need to show it is bounded. However this is also true; since you assume $\lim_{n\to\infty} \sum_{k=1}^na_k$ exists, it is bounded, as there is some $N\in\mathbb{N}$ such that $$\sum_{k=1}^ma_k<1+\lim_{n\to\infty} \sum_{k=1}^na_k $$ for $m\geq N$.