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Consider a sequence of real numbers $\{a_i(n)\}_{i=1}^{n}$, with $|a_i(n)|<1$. I write $a_i(n)$ because each element of the sequence is a function of the total number $n$ of elements. Assume that it can be proven that the infinite series is absolutely convergent

$$\sum_{i=1}^{\infty} |a_i(n)| < \infty \tag{1}$$

and that

$$\sup_i |a_i(n)|\to 0,\;\; \text{as}\;\;n\to \infty \tag{2}$$

The supremum in eq. $(2)$ is calculated over all elements of the sequence of absolute values. So here all the elements of the sequence are sandwiched to zero.

Set $S_n = \sum_{i=1}^{n} |a_i(n)|$. Are the previous results enough to obtain that

$$\lim_{n \to \infty} S_n = 0\;\;? \tag{3}$$

It looks like the infinite series tends to the indeterminate form $0\cdot \infty$ in which case $(3)$ has to be examined per case and so it does not hold at this level of generality. Still, I may have been trapped by false intuition that cries "zero", and moreover I wasn't able to construct a non-zero counter example, neither was I able to prove $(3)$.

Any ideas? If indeed $(3)$ does not hold in general, are there any known additional general conditions that would lead to $(3)$?

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    Will you clarify what the index/variable $n$ is used for here? It appears $a_i$ is a function of $n$ and you are saying that for all $n\in\Bbb N$, the infinite series is absolutely convergent. Sorry if I'm just missing something simple. – 2017-01-25
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    @Clayton Clarified. $n$ is the total number of elements in the sequence, and yes, each element depends on $n$ (this is why the supremum goes to zero). – 2017-01-25
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    Does something like $a_i(n)=\frac{1}{n+i^2}$ suffice as a counterexample? It's absolutely convergent, and as $n\to\infty$, $a_i(n)\to0$ (hence the supremum is $0$). – 2017-01-25
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    @Nameless No, $n$ is the number of elements in the sequence. – 2017-01-25
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    Do you want a limit as $n \to \infty$ in (3)? – 2017-01-25
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    @AlecosPapadopoulos, your sequence is $\{a_i(n)\}_{i}^{\infty}$, indexed $i$ and is an *infinite* sequence. What you mean the number of elements in the sequence? – 2017-01-25
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    @UmbertoP. Yes, that's the limit I want to evaluate and I wonder whether it equals zero or not – 2017-01-25
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    The question should be edited to reflect that since (as evidenced by all these comments) it is confusing what was meant. – 2017-01-25

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Take $a_i(n)$ to be the sequence $\dfrac 1n, \dfrac 1n,\ldots,\dfrac 1n,0,0,\ldots$ where the number of nonzero terms is $n$. Then:

1) $S_n = \displaystyle \sum_{i=1}^n |a_i(n)| =\sum_{i=1}^\infty |a_i(n)| = 1$

2) $\displaystyle \sup_i |a_i(n)| = \frac 1n \to 0$

but 3) $\displaystyle \lim_{n \to \infty} \sum_{i=1}^\infty |a_i(n)| = 1.$

You need some form of dominated convergence.

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    I love this example. it's more pertinent that you might suspect. Thanks! – 2017-01-25
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    In the first line, the zeros should not, _could not_ be there -and they are not needed anyway. – 2017-01-25