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.

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

  • $\begingroup$ Thank you, but why did you say that $N$ is necessarily greater than $n$? $\endgroup$ – Steve Jul 14 '20 at 11:25
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    $\begingroup$ @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 $n < N$. $\endgroup$ – Air Conditioner Jul 14 '20 at 12:15
  • $\begingroup$ This is just what I want to know. Thank you all. $\endgroup$ – Steve Jul 14 '20 at 15:25

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