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Suppose we are given a sequence of nonnegative real numbers:

$a_1\geq a_2\geq a_3\geq\dots\geq 0$

such that

$\lim_{n \to \infty} a_n=0.$

Assume that

$\lim_{k\to\infty} (a_{n_k}\ln n_k)=0$

for some sequence $(n_k)_{k=0}^\infty$ of natural numbers.

Is it true that

$\lim_{n\to\infty} (a_{n}\ln n)=0 $

also holds?

Keep in mind that the sequence $a_n$ need not to be strictly monotonic.

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    Note also that requiring $a_n$ to be strictly monotonic changes nothing, as the examples Andre and I give can be modified to have $a_n$ decrease very slightly between $a_{n_k}$ and $a_{n_{k+1}}$ without changing the result.2012-08-27

1 Answers 1

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No. Let $n_1=2$ and $n_{k+1}=\lceil n_k^{\ln n_k}\rceil+1$. Define $a_{n_k}=\frac{1}{(\ln n_k)^2}$, and extend this to all indices by letting $a_n=a_{n_k}$ for the greatest $n_k$ less than $n$. Clearly $\lim\limits_{n\to\infty}{a_n}=0$ and $\lim\limits_{k\to\infty} a_{n_k}\ln n_k=0$. However, for $n=n_{k+1}-1$ we have $a_n\ln n=\frac{1}{(\ln n_k)^2}\ln \lceil n_k^{\ln n_k}\rceil\geq \frac{1}{(\ln n_k)^2}\ln n_k^{\ln n_k}=1$ and thus $\lim\limits_{n\to\infty} a_n\ln n$ does not exist.

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    Thank you for your answer.2012-08-27