Thread: Problem 36
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Old September 11th, 2007, 04:33 AM
albi albi is offline
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Hm, isn't it too early to post the solution, and make the other people have no fun?

I would like to do this more elegant simply taking \epsilon > 0. From Cauchy condition we have (as you noticed): (k-l)a_l < a_l + a_{l+1} + \ldots + a_k < \epsilon. Now we simply take k = 2n and l = n to get na_n < \epsilon.

That means this positive sequence converges to 0.