# limit of the sum ∑Ni=1aN(i) \sum_{i=1}^N a_N(i)

I have that strange sum:
N∑i=1aN(i)
\sum_{i=1}^N a_N(i)

for which I want to compute the limit as N→∞N\to \infty, except that I have as an additional condition limN→∞aN(i)=0\lim_{N\to \infty}a_N(i)=0.

I have that bizzare intuition that this serie should converge to zero except that I can’t figure how ? or maybe my intuition is wrong !! What do you think ?

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Your notation is kind of strange. a_N(i)a_N(i) is a function of both NN and ii?
– D_S
2 days ago

yes, it’s a function that depends on both ii and NN.
– houda
2 days ago

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2

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The series need not converge to zero, if all you know is that \lim_{N\to\infty}a_N(i)=0\lim_{N\to\infty}a_N(i)=0. Take, for instance, the degenerate case

a_N(i) = \frac1N,\qquad\text{for all $i=1,\ldots N$}.

a_N(i) = \frac1N,\qquad\text{for all $i=1,\ldots N$}.

Then \sum_{i=1}^N a_N(i)\sum_{i=1}^N a_N(i) is equal to 11 for every NN.

yes you’re right ! I’ve missed that :p thanks.
– houda
2 days ago

In order to make this conclusion you need additional conditions. A popular one is that there is some sequence b_i \ge 0b_i \ge 0 such that \sum_{i=1}^\infty b_i < \infty\sum_{i=1}^\infty b_i < \infty and |a_N(i)| \le b_i|a_N(i)| \le b_i for all ii and NN.      so under this condition I can conclude that the limit of \sum_{i=1}^N a_N(i)\sum_{i=1}^N a_N(i) is zero as N\to \inftyN\to \infty ? if it's okay can you give a reference as I want to see the proof 🙂 – houda 2 days ago      It's not hard to prove on your own. You need to put two results together: first, if \sum\limits_{i=1}^{\infty} |a_i|\sum\limits_{i=1}^{\infty} |a_i| converges, then so does \sum\limits_{i=1}^{\infty} a_i\sum\limits_{i=1}^{\infty} a_i. Second, if 0 \leq a_i \leq b_i0 \leq a_i \leq b_i for all ii, and if \sum\limits_{i=1}^{\infty} b_i\sum\limits_{i=1}^{\infty} b_i converges, then so does \sum\limits_{i=1}^{\infty} a_i\sum\limits_{i=1}^{\infty} a_i. This is because 0 \leq \sum\limits_{i=1}^N a_i \leq \sum\limits_{i=1}^N b_i \leq \sum\limits_{i=1}^{\infty} b_i < \infty0 \leq \sum\limits_{i=1}^N a_i \leq \sum\limits_{i=1}^N b_i \leq \sum\limits_{i=1}^{\infty} b_i < \infty for all NN. Now, \sum\limits_{i=1}^N a_i\sum\limits_{i=1}^N a_i is a bounded increasing sequence, so it must converge. – D_S 2 days ago      It's a special case of the Lebesgue dominated convergence theorem. – Robert Israel 21 hours ago