# Absolutely convergent series, trying to find a counterexample of a statement.

If $\sum a_k$ converges absolutely, then $|a_k|<\frac{1}{k}$ for all sufficiently large k.

I'm trying to give a proof or a counterexample about the above statement, but I'm not really sure where to start. I think it's false, but don't know how to go about finding a counterexample. Any help would be much appreciated.

Thanks

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Do you mean $|a_k| < \frac {1}{k}$? –  Calvin Lin Dec 30 '12 at 0:10
Yep I did, thanks –  user51327 Dec 30 '12 at 1:10

## 1 Answer

HINT: The idea is to have infinitely many terms $a_k$ such that $a_k\ge\frac1k$, but to spread them out very thinly. Try letting

$$a_k=\begin{cases} \frac1k,&\text{if }k\text{ is a perfect square}\\\\ 0,&\text{otherwise}\;. \end{cases}$$

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@Jonas: It means that I didn’t include a proof that the example works. I should perhaps have said BIG HINT. –  Brian M. Scott Dec 30 '12 at 0:03
I still don't understand how you came up with the example? And why do you want to spread them out thinly? –  user51327 Dec 30 '12 at 1:10
@user51897 The intuition is that $\int \frac {1}{k} dk = \ln k$, so when evaluated from $n=k$ to infinity, has the value $\infty$. This means that the terms need not converge. As such, we need it to be less frequent than that. Brian's intuition is that $\int \frac {1}{k^2} dk = \frac {1}{k}$, which when evaluated as a definite integral, is finite. –  Calvin Lin Dec 30 '12 at 1:16
@user51897: Because the harmonic series diverges, so you can't have "too many" terms such that $a_k\geq \frac{1}{k}$. If you are going to have infinitely many terms such that $a_k\geq \frac1{k}$, they have to be chosen carefully from a set such that those reciprocals do converge. The sum of the reciprocals of the squares converges. Another example would be the sum of the reciprocals of the powers of $2$. –  Jonas Meyer Dec 30 '12 at 1:17
Ok, so I think I finally understand that. How do you start a proof about an absolutely convergent series? –  user51327 Dec 30 '12 at 16:54