# $\lim_{n\to\infty}n^{n\cdot a_{n}}=1$ as a sufficient condition for $\sum a_{n}<\infty$

[Not HW]

Let $\left(a_{n}\right)_{n=1}^{\infty}$ a sequence of real numbers such that, for all $n\in\mathbb{N}$ , $0<a_{n}\leq\frac{1}{2}$ .

Why the following statements don't imply the convergence of $\sum_{n=1}^{\infty}a_{n}$ :

$$\lim_{n\to\infty}n^{n\cdot a_{n}}=1$$

and the second one:

$$\frac{a_{n+1}}{a_{n}}<1-\frac{1}{n+1}$$

For the second one, since the ratio test calls for the $\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$ to be strictly less than 1, it's obvious that the above statement does not imply convergence, but I didn't find a counterexample neither.

At the exam, I did mark statement 1 as a sufficient condition for convergence, but it's not true. So we need to find $a_{n}$ such that:

• $\lim_{n\to\infty}n^{n\cdot a_{n}}=1$ $\iff$ $\lim_{n\to\infty}n\cdot a_{n}\ln n=0$
• $0<a_{n}\leq\frac{1}{2}$
• $\sum_{n=1}^{\infty}a_{n}$ diverges.

I thought of the following sequence:

$$a_{n}=\frac{1}{2},\frac{1}{2},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{8},\frac{1}{16}\dots$$

$\sum a_{n}$ clearly diverges but I'm not sure about $\lim_{n\to\infty}n^{n\cdot a_{n}}$ .

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Take $a_n = 1/(n \ln n \ln \ln n)$. –  Rijul Saini Aug 1 '12 at 5:24
@RijulSaini I guess that's a standard counterexample then (didn't see your comment when answering). –  Alex Becker Aug 1 '12 at 5:36
Great! Thank you! –  Amihai Zivan Aug 1 '12 at 5:38

Your reasoning for the second statement is invalid. While the given condition that $$\tag{1}{a_{n+1}\over a_n}<1-{1\over n+1}$$ is allowing the possibility that the limit of $a_{n+1}\over a_n$ is 1, the Ratio Test does not imply anything concerning divergence of the corresponding series in this case.

Note that condition $(1)$ is not allowing $a_{n+1}\over a_n$ to approach 1 "too fast". In fact, if we demand the stronger condition that $$\tag{2}{a_{n+1}\over a_n}<1-{p\over n+1}$$ for some $p>1$ and sufficiently large $n$, then the corresponding series converges. This is known as Raabe's Test (one version of this is excersise 28 here).

With $p=1$, as in your condition, the series may diverge: for example, take take $a_n={1\over n\ln n}$, $n>1$. Then for $n$ sufficiently large, ${a_{n+1}\over a_n}= {n\ln n\over (n+1)\ln (n+1)}<{n\over n+1}=1-{1\over n+1}$. The integral test will show that $\sum{1\over n\ln n}$ is divergent.

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Taking logarithms, the first condition is equivalent to the fact that $\lim\limits_{n\to\infty}a_nn\log n=0$. This is not enough to guarantee that $\sum\limits_na_n$ converges, consider for example $a_n=1/(n\log n\log\log n)$ for every $n\geqslant3$.

The second condition is equivalent to the fact that the sequence with general term $na_n$ is strictly decreasing. This is not enough to guarantee that $\sum\limits_na_n$ converges, even with the further condition that $na_n\to0$, consider for example $a_n=1/(n\log n)$ for every $n\geqslant2$.

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Your example does not work, in fact $\lim\limits_{n\to\infty} n^{n\cdot a_n}=\infty$ in its case, as we have $\frac{1}{2n}\leq a_n$ so $n^{n\cdot a_n}\geq \sqrt{n}\to \infty$. An example which does work is $a_n=\frac{1}{n\ln(n)\ln(\ln(n))}$ as $$\lim_{n\to\infty}n^{n\cdot a_n}=\lim_{n\to\infty} n^{\frac{1}{\ln(n)\ln(\ln(n))}}=\lim_{n\to\infty}(e^{\ln n})^{\frac{1}{\ln(n)\ln(\ln(n))}}=\lim_{n\to\infty}e^{1/\ln(\ln(n))}=1$$ while the sum diverges, according to WolframAlpha.

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And according to Cauchy's condensation test.. –  Amihai Zivan Aug 1 '12 at 5:38
And by the integral test-the derivative of log composed with itself $n$ times is one over the product of $n,$ $\log(n)$, all the way up to log composed with itself $n-1$ times. In particular this series is the derivative of $log(log(log(n)))$. –  Kevin Carlson Aug 1 '12 at 5:40

$\lim_{n\to\infty}n^{na_n}=1\implies \lim_{n\to\infty}na_n\log n=0\implies a_n\to 0$ more rapidly than $n\log n\to\infty\implies$ order of $a_n$ is somewhat "lesser"(don't know if this is the right term) than $n\log n$ ,for example let $a_n=\frac{1}{n^2}$,then $\sum_0^{\infty}a_n$ converges to $\frac{\pi^2}{6}$ while if you take $a_n=\frac{1}{n(\log n)(\log(\log n))}$ then the series diverges. So, both of the above examples of $a_n$ above satisfy $\lim_{n\to\infty}n^{na_n}=1$ but in one case the series $\sum_0^{\infty}a_n$ converges and diverges in the other one.

Now for the second part, if $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ (this is your case), then the ratio test fails to tell anything about convergence or divergence of the series. For example, take $a_n=\frac{1}{n}$, and $a_n=\frac{1}{n^2}$; ratio test fails( i.e. $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$) for both of these but in first case series diverges while converges in the second one.

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$\sum \frac{1}{n(\ln n)^2}$ converges. –  Amihai Zivan Aug 1 '12 at 8:04