# A sequence, whose all possible finite sub sums are bounded, converges

I have the following problem I can't solve. Your help would be most appreciated!

A sequence $(a_n)$ fulfills the following condition:
the set of all possible finite sub sums of $(a_n)$ is bounded. that is there exists an $M$ such that $|\sum_k a_{n_k}| < M$ ($k$ is taken from any sub sequence of the natural numbers).
Prove that the series $\sum_n a_n$ converges absolutely.

• Use an induction and add one term – abiessu May 31 '17 at 13:24

For a finite subset $F$ of the natural numbers, let $$s_F=\sum_{n\in F}a_n$$ The condition on the sequence can be rewritten, perhaps more clearly

There exists $M>0$ such that, for every finite subset $F$ of the natural numbers, $|s_F|<M$.

Set now $\hat{s}_F=\sum_{n\in F}|a_n|$. Then prove that, for each $F$, you have $$0\le \hat{s}_F <2M$$ (hint: divide between the positive and negative elements $a_n$, for $n\in F$).

Then the series $$\sum_{n=1}^{\infty}|a_n|$$ converges because the partial sums make a bounded and increasing sequence.

Let us define two sequences :

• $a^{+}_{n}=a_n$ if $a_n\geq 0$ and $0$ otherwise.
• $a^{-}_{n}=a_n$ if $a_n\leq 0$ and $0$ otherwise.

Then we have $|a_n|=a_n^{+}-a_n^{-}$. Moreover, $a_n^{+ }\geq 0$ for all $n$, and $a_n^{- }\leq 0$ for all $n$ ; and since any subsequence of $a_n^{\pm}$ is also a subsequence of $a_n$, the partial sums of $a_n^{\pm}$ are bounded, therefore $\sum_n a_n^{\pm}$ converges. Now $\sum_n |a_n|$ is the sum of two convergent series, and is therefore convergent.