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Given a positive sequence $\{a_n\}$, if $$\lim _{n \rightarrow \infty} \ln n \cdot\left(\frac{a_n}{a_{n+1}}-1\right)=\lambda>0,$$ prove that $\displaystyle\lim _{n \rightarrow \infty} n a_n=0$.

Although the given condition appears to have a similar form to the Raabe's test, there are significant differences between them which make it difficult to establish the proof.

In this case, we can use the Raabe's test to show that the sequence is approaching zero and then attempt to apply the Stolz-Cesaro theorem. Specifically, we have $$\lim _{n \rightarrow \infty} n a_n=\lim _{n \rightarrow \infty} \frac{a_{n+1}-a_n}{\frac{1}{n+1}-\frac{1}{n}}=-n(n+1)(a_{n+1}-a_n).$$ However, there does not seem to be an obvious way forward from here.

Any help or comments on this matter would be greatly appreciated to further advance the proof.

unicornki
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1 Answers1

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$\frac{a_n}{a_{n+1}}-1$ must be positive for sufficiently large $n$, so that $(a_n)_{n \ge N}$ is decreasing for some $N$. Also $$ n \cdot\left(\frac{a_n}{a_{n+1}}-1\right) = \frac{n}{\ln n}\cdot \ln n \cdot\left(\frac{a_n}{a_{n+1}}-1\right) \ge \frac{n}{\ln n} \frac 12 \lambda > 2 $$ for sufficiently large $n$, so that $\sum_{n=1}^\infty a_n$ is convergent by Raabe's test.

Finally use that

If $\{a_n\}$ is a nonincreasing sequence of positive real numbers such that $\sum_n a_n$ converges, then $\lim_{n \rightarrow \infty} n a_n = 0$.

which is proven e.g. here: If $(a_n)\subset[0,\infty)$ is non-increasing and $\sum_{n=1}^\infty a_n<\infty$, then $\lim_{n\to\infty}{n a_n} = 0$

Martin R
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