# How to prove that if $\sum|a_j|<\infty$, then $\sum|a_j|\log(|a_j|^{-1})<\infty$

I would like to prove the following statement about series.

Suppose that $\{a_j:j\ge1\}$ is a real sequence and $a_j\ne0$ for each $j\ge1$. If the series $\sum_{j=1}^\infty|a_j|<\infty$, then the series $$\sum_{j=1}^\infty|a_j|\log(|a_j|^{-1})<\infty.$$

It is not difficult to show that this statement is true for some particular sequences $\{a_j:j\ge1\}$. For example, suppose that $a_j=j^{-2}$ and use the integral test for convergence. But I have no idea how to prove the statement for a general sequence $\{a_j:j\ge1\}$.

Any help would be much appreciated!

• Your general term is negative for big values of $j$, are you sure that the series isn't $\sum |a_j||\log(1/|a_j|)|$ ? By the way, where did you find this exercise ? – user37238 Oct 30 '13 at 9:00
• @user37238 I think that the general term is positive for big values of $j$. Since $|a_j|\to0$, we have that $|a_j|^{-1}\to\infty$ and $\log(x)>0$ for $x>1$. The general term might be negative for small values of $j$. Could you give me some details why the general term is negative for big values of $j$? – Cm7F7Bb Oct 30 '13 at 12:34
• @user37238 This exercise occured to me while solving some other problems. I'm not sure if it is stated somewhere. The statement might even be false, but I can't find a counterexample. – Cm7F7Bb Oct 30 '13 at 13:04

Try $a_j=1/(j(\log j)^c)$ for every $j\geqslant3$, then $\log(1/a_j)\sim\log j$ hence $\sum\limits_ja_j$ converges exactly when $c\gt1$ and $\sum\limits_ja_j\log(1/a_j)$ converges exactly when $c\gt2$.

Thus, any sequence defined by $a_j=1/(j(\log j)^c)$ for every $j\geqslant3$, for some $1\lt c\leqslant2$, disproves your claim.

$\sum_{j=1}^\infty|a_j|\log(|a_j|^{-1})=-\sum_{j=1}^\infty|a_j|\log(|a_j|)$.

Now $\log(|a_j|)\leq |a_j|$ and thus $\sum_{j=1}^\infty|a_j|\log(|a_j|)\leq \sum_{j=1}^\infty |a_j|^2<\infty$
because $\sum_{j=1}^\infty|a_j|<\infty$ we have that $|a_j|\to 0$ thus there is a $k\in \Bbb N$:$|a_j|<1 \forall j\geq k$. So $|a_j|^2\leq |a_j|<1 \forall j\geq k$.

Thus $\sum_{j=1}^\infty |a_j|^2\leq \sum_{j=1}^\infty|a_j|<\infty$.

So $-\sum_{j=1}^\infty|a_j|\log(|a_j|)<\infty$.

• Note $\sum|a_j|\log(|a_j|)$ is a series with almost all terms negative. You just proved this series bounded above, but boundedness below is not resolved here. – Norbert Oct 30 '13 at 11:51