# Is $\sum\limits_{n=2}^{\infty} \frac{(-1)^n n^2}{( \ln(n))^n}$ divergent?

The series in question is $\sum\limits_{n=2}^{\infty}\frac{(-1)^n n^2}{( \ln(n))^n}$ I think this diverges. So using the divergence test, I am trying to show that limit of the general term of the series is not 0. By a few algebraic manipulations we get :

• $\lim (-1)^n(n^2)( \ln(n))^{-n}$
• $\lim (-1)^n(n)/ \ln(n)$

I am not sure how to proceed here. If you get rid of the alternating, the limit is $\infty$. However, I am not sure how to manipulate this to show that (I suspect) the limit does not exist.

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You should fix the $\TeX$ formulas, so we know what you really mean. – Mariano Suárez-Alvarez Nov 1 '10 at 4:05
Due to formatting problems, I can't tell what series you are trying to sum. In particular I don't know what to do with the -n at the end. But Liebnitz' test may apply-if the terms decrease monotonically to zero, it converges. – Ross Millikan Nov 1 '10 at 4:06
You are not trying to show the "lim of the series is not $0$"; you are trying to show the limit of the general term of the series is not zero (that's how you would apply the Divergence Test). The divergence test is about the terms, not about the series. – Arturo Magidin Nov 1 '10 at 4:16

Applying the root test you can see that the series converges.

$$L = \lim_{n\to\infty} |a_{n}|^{1/n} .$$

If $L<1$, then the series is absolutely convergent.

If $L>1$ the series is divergent.

If $L=1$ then the test is inconclusive. However, if $|a_{n}|^{1/n}\geq 1$ for infinitely many distinct values of n, then the series $\sum_{n=1}^{\infty} a_{n}$ diverges.

In your series, you have

$$L = \lim_{n\to\infty}\bigg| \frac{(-1)^{n}n^{2}}{\text{ln}^{n}(n)}\bigg|^{\frac{1}{n}}$$

Then $L=\lim \limits_{n\to\infty}\bigg|\displaystyle\frac{(-1)n^\frac{2}{n}}{\text{ln}(n)}\bigg|=0$ (because $\lim \limits_{n\to\infty}\ln(n)=\infty$).

Therefore, $L<1$ and the series convergences.

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This is a better approach than mine. It is natural considering the $n^{th}$ power there, and the result is stronger because you have absolute convergence. – Jonas Meyer Nov 1 '10 at 4:52
In our textbook it has slightly different conclusions when applying the ratio or root test. – fdart17 Nov 1 '10 at 13:42
if $L < 1$, then then series converges (but not necessarily absolutely) if $L > 1$, then the series diverges if $L = 0$ the test is inconclusive. Thanks – fdart17 Nov 1 '10 at 13:42
@Fdart17: Check carefully the hypothesis of the root test in your book; if it is restricted to series with positive terms (a common restriction), then the result for absolute convergence will follow by taking the absolute value. – Arturo Magidin Nov 1 '10 at 14:17
"If $L=1$ the test is inconclusive (except "...when the limit approaches strictly from above then the series diverges")." You mean the series is approaches from greater than 1? – fdart17 Nov 4 '10 at 1:28

$$\left| \frac{(-1)^n n^2}{( \ln(n))^n} \right| = \frac{n^2}{( \ln(n))^n} < \frac{n^2}{2^n} \, \mbox{ for all } n>e^2 \,.$$

Since $\sum \frac{n^2}{2^n}$ is convergent, you get that your series is absolutely convergent.

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+1. I am surprised no quantitative answer (like this one) was posted earlier. – Srivatsan Dec 28 '11 at 0:35

The series converges by the alternating series test. Take logs of the absolute values of the terms to get $2\ln(n)-n\ln(\ln(n))$. Consider the function $f(x)=2\ln(x)-x\ln(\ln(x))$, with derivative $f'(x)=\frac{2}{x}-\ln(\ln(x))-\frac{1}{\ln(x)}$. Since $f$ goes to $-\infty$ and $f'$ is eventually always negative, continuity and monotonicity of the exponential function tell us that the terms of your sequence have absolute value eventually decreasing to 0, so that the alternating series test applies.

The second expression you give for the limit does not go to zero, and it is not clear where it came from.

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If $a_n$ is a sequence, then $\lim_{n\to\infty}a_n = 0$ if and only if $\lim_{n\to\infty} |a_n| = 0$. So if the absolute value of the terms does not go to zero, then the terms don't go to zero.

In other words, if you know that $\lim \frac{n}{\ln (n)}\neq 0$, then it follows that $\lim\frac{(-1)^n n}{\ln n}\neq 0$. (because the former is the limit of the absolute values of the latter). So if your reduction is accurate, and your proof that the limit without the $(-1)^n$ factor is not zero is accurate, then you are done.

That said, I'm not sure how you manage to get to that limit. If you had $(\ln(n^n))^{-1}$, then I would agree; but how do you manage to get rid of the exponent and cancel an $n$ if you have $\frac{(-1)^n n^2}{(\ln n)^n} \text{?}$

Added: Well, looks like the beans have been spilt; your "few algebraic manipulations" were incorrect. Remember that while $\ln(a^k) = k\ln(a)$, it is not true that $(\ln a)^k = k\ln a$. $\ln(a^k)$ is the logarithm of the $k$th power of $a$; $(\ln a)^k$ is the $k$th power of the logarithm of $a$; they are usually different.

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You want to prove that $n^2 (\ln (n))^{-n}$ is monotonically decreasing and use Leibniz' test. As long as $\ln (n)$ is greater than $1$, the exponential will dominate the polynomial. So pick your favorite starting place and fix $\ln (n)$ to $\ln(a)$ for some $a$ and argue that the actual term is smaller in absolute value than this. Now show that the version with $\ln(a)$ also goes to zero.

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HINT: Consider $$u_n=\left| \frac{(-1)^n n^2}{( \ln(n))^n} \right|$$ and then apply the ratio test / root test.

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