# Prove that $\sum {{a_n}}$ converges iff the sequence of partial sums is bounded where $a_n\geq 0$

Let (${a_n}$) be a sequence of nonnegative real numbers. Prove that $\sum {{a_n}}$ converges iff the sequence of partial sums is bounded.

• Perhaps you want the $a_n$ to be non-negative? Then your result follows from the monotone convergence theorem. – Kenny Wong Apr 26 '17 at 23:22
• #MathematicsStudent112 #Kenny Wong I'm sorry. I mean nonnegative – Itsnhantransitive Apr 26 '17 at 23:23

Assuming you mean nonnegative sequence of real numbers (it is very false otherwise), here's a hint: Increasing sequences bounded above converge...

• Ok so I just need to show if it is bounded above! – Itsnhantransitive Apr 26 '17 at 23:25
• @NTT bounded$\implies$ bounded above – qbert Apr 26 '17 at 23:26

A basic theorem from real analysis is that bounded, monotone sequences converge. So, in this case, since the $a_n$ are nonnegative, the sequence $\{S_m= \sum_{n=1}^{m} a_n\}$ of partial sums is nondecreasing. Also by assumption $S_m$ is bounded. So applying the theorem we get that $\lim_{m\rightarrow\infty} S_m:=\sum_{n=1}^{\infty}a_n$ converges. Conversely, if the partial sums are not bounded (above), then for all $M>0$, there exists an $N_M\in\mathbb{N}$ such that for all $m\ge N_M$, $$S_m\ge S_{N_M}>M.$$ This shows that for any real $M$ that you can find an $0<\epsilon_M<S_{N_M}-M$ such that $S_m\not\in (M-\epsilon_M, M+\epsilon_M)$ for all $m>N_M$. By definition, $S_m$ cannot converge. This proves the contrapositive.

To prove the theorem I mentioned, since $\{S_m\}$ is nondecreasing (WLOG, else replace sup with inf) and bounded, and since $\mathbb{R}$ has the LUBP, $L:=\sup\{S_m\}$ exists. By definition of the supremum, for all $\epsilon>0$, there exists an $N$ such that $L-\epsilon<S_N\le L$. Since the sequence of partial sums is nondecreasing, for all $m\ge N$, we have $$L-\epsilon<S_N\le S_m\le L,$$ so that for all $m\ge N$, $\vert L-S_m\vert <\epsilon$. By definition of the limit, $L=\lim_{m\rightarrow\infty} S_m$.

If you know measure theory, this entire exercise is just an application of the Monotone Convergence Theorem.

• Oh I see, I found a theorem in my book that is called Cauchy Criterion for Series. Can I use that too? – Itsnhantransitive Apr 27 '17 at 0:48
• @NTT In $\mathbb{R}$ a series is convergent iff it's Cauchy. So I suppose you could prove the partial sums are Cauchy iff the partial sums are bounded, but I don't see an advantage for that. – Satana Apr 27 '17 at 5:42