# 2015 微積分一期中考第一部分證明題
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**1.**(6 points) Show that
$$\sum\limits_{n=1}^\infty{a_n}=\sup\limits_{n_1<\dots<n_k}{\{a_{n_1}+\dots+a_{n_k}\}}$$ if $a_n\ge 0$ for all $n$. (Here we allow any side to be $\infty$.)
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**Proof.** Let $(a_n)_{n\in\mathbb{N}}$ be a sequence in $\mathbb{R}$ so that $a_n\ge 0$ for all $n$ and let $s_n:=a_1+\dots+a_n$ denote its $n$-th partial sum. Since $s_{n+1}=s_n+a_{n+1}\ge s_n$ , $(s_n)_{n\in\mathbb{N}}$ is monotone increasing. By the monotone convergence theorem, $\sum\limits_{n=1}^\infty{a_n}=\lim\limits_{n\to\infty}{s_n}=\sup\limits_{n\in\mathbb{N}}{s_n}$. Since
$$a_{n_1}+\dots+a_{n_k}\le\sum\limits_{n=1}^{n_k}=s_{n_k}\le\sup\limits_{n\in\mathbb{N}}{s_n}=\sum\limits_{n=1}^\infty{a_n},$$ for all $n_1<\dots<n_k$, we have $$\sum\limits_{n=1}^\infty{a_n}\ge\sup\limits_{n_1<\dots<n_k}{\{a_{n_1}+\dots+a_{n_k}\}}.$$ Since for every $n$
$$s_n=a_{n_1}+\dots+a_{n_k}\le \sup\limits_{n_1<\dots<n_k}{\{a_{n_1}+\dots+a_{n_k}\}}$$if we choose $n_i=i$ for all $i$, we have
$$\sum\limits_{n=1}^\infty{a_n}=\sup\limits_{n\in\mathbb{N}}{a_n}\le\sup\limits_{n_1<\dots<n_k}{\{a_{n_1}+\dots+a_{n_k}\}}.$$ Hence it follows that $$\sum\limits_{n=1}^\infty{a_n}=\sup\limits_{n_1<\dots<n_k}{\{a_{n_1}+\dots+a_{n_k}\}}.$$