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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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Is the sequence $\{n_k\}$ known to be unbounded? –  Sasha Aug 27 '12 at 18:57
    
Of course. $a_{n_k}$ is a proper subsequence of $a_n$. –  Classical Analyst Aug 27 '12 at 19:01
    
This fails in general, but should work if $n_k$ grows slowly enough. –  Alex Becker Aug 27 '12 at 19:14
1  
For simplicity look instead at logs to the base $2$. For $k\ge 1$ and $2^k \le n \le 2^{k^3}$, let $a_n=\frac{1}{k^2}$. –  André Nicolas Aug 27 '12 at 19:15
    
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. –  Alex Becker Aug 27 '12 at 19:20

1 Answer 1

up vote 4 down vote accepted

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. –  Classical Analyst Aug 27 '12 at 20:16

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