# Prove that each partial sum of a convergent series of non-negative terms cannot exceed the sum of the series by elementary calculus

Let $$\sum_{n=1}^\infty a_n$$ be a convergent series of non-negative terms. And its sum is denoted by $$S$$. Let $$S_k$$ be the $$k$$-th partial sum of the series. I would like to prove that $$\forall k\in\mathbb N, S_k\leq S$$ rigorously but without appealing to real analysis. In a real analysis course, one can learn that $$S=\sup\{S_k:k\in\mathbb N\}$$ since $$S_k\nearrow S$$. This enables us to conclude the result immediately. But what if we can only use calculus(e.g. Thomas' calculus book)? Is there anything we can do? Thank you.

• Not sure what you mean by "without real analysis but with calculus". What are you allowed to use here? – Mark Jul 14 '20 at 10:00
• No $\inf$, $\sup$, point-set topology, continuum of real numbers, etc. You can use any calculus book such as Stewart, Thomas, Salas, Larson, – Steve Jul 14 '20 at 11:31

Suppose $$S_n\gt S$$. Let $$\epsilon=(S_n-S)/2$$. By definition of $$S$$, there is an integer $$N$$ such that $$|S_k-S|\lt\epsilon$$ for all $$k\ge N$$. Since $$|S_n-S|=S_n-S=2\epsilon\gt\epsilon$$, $$N$$ is necessarily greater than $$n$$. But
$$S_N=S_n+a_{n+1}+\cdots+a_N\ge S_n$$
since $$a_i\ge0$$ for all $$i$$, which implies $$|S_N-S|=(S_N-S_n)+(S_n-S)\ge S_n-S\gt\epsilon$$, a contradiction.
• Thank you, but why did you say that $N$ is necessarily greater than $n$? – Steve Jul 14 '20 at 11:25
• @Steve For all $k \ge N$, we have $|S_k-S|< \epsilon$. Thus, if $n \ge N$, $|S_n-S|$ would be less than $\epsilon$. However, we defined $\epsilon$ so that $|S_n-S|=2 \epsilon > \epsilon.$ Since $|S_n-S| \not< \epsilon$, we must have $n < N$. – Air Conditioner Jul 14 '20 at 12:15