# How to deal with (ln(x))^p in an Alternating Series Test

1. May 11, 2015

### Jrb4935

1. The problem statement, all variables and given/known data
Determine all values of P for which the series ∑((-1)^(n-1))((ln(x))^p)/(5n) is convergent, expressing your answer in interval notation (Problem is shown in attached picture).

2. Relevant equations
• Alternating Series Test: If {a_n} is positive and decreasing, and if the lim as n→∞{a_n}=0, then the sum ∑((-1)^(n-1))(b_n) converges.
• In this problem, {a_n} is ((ln(x))^p)/(5n), and that is what is supposed to be evaluated for it's limit to test for convergence.

3. The attempt at a solution
The concept of the alternating series test and the method of using it is not my problem, I have a fairly good grasp on that. The problem is that I am clueless on what to do with ((ln(n))^p, as I have never been taught what you can do with that in any situation at all, and after hours of searching the internet/asking around, I still could not find any ln(n) laws that have to do with the entire ln(n) function being raised to a power rather than just whats inside the parenthesis being raised to a power.

The online homework in question has already been submitted, and the answers are given. But, this is where probably my biggest problem lies. The explanation says that the limit of the given {a_n} is=0 for any value of P (-∞,∞). But just using simple logic, if you choose a value of P that is large like 65 and evaluate the limit with n equal to say 20, then the value of the numerator would be much larger than the value of the denominator, meaning the limit would be infinity and not 0...

Could really use some clarification for these 2 things, this is for integral calculus 2 and I am trying to clear up this confusion while preparing for the final exam.

Thanks for your time and effort

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2. May 11, 2015

### RUber

I think the question is asking if there is a fixed p in $(-\infty, \infty)$ such that the series does not converge to zero.
Focus on the rate of growth in the terms, and notice that they are alternating signs, so if the rate of growth is slowing, then the grouped sum
$\sum_{k=1}^\infty \frac {(\ln (2(k+1)) )^p}{5(2k+1)}-\frac {(\ln (2k) )^65}{10k}$ would satisfy the condition of positive and decreasing.
The shape of the function grows quickly for large p and then comes back down as n gets large.
See the attached plot for an example with p = 7. The larger p gets, the more extreme the values might be, but the overall shape will still tend back to zero as n gets really large.

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3. May 11, 2015

### Ray Vickson

If you think about it you should realize that the alternating series test (AST) does not need all the $a_n$ positive and decreasing; it just needs $a_n > 0$ decreasing to 0 for all $n \geq N$ for some finite $N$. For example, the first 20 $a_n$ values could be increasing, but if they are positive and decreasing to zero for $n \geq 21$ then the 'tail' series $\sum_{n=21}^{\infty} (-1)^n a_n = \sum_{k=1}^{\infty} (-1)^{20+k} b_k$ satisfies the AST; here, $b_k = a_{20+k}$. This happens when $p = 3$ for example.

So, the series will converge if $a(n) = \log(n)^p / n$ decreases eventually and has a limit of 0. What happens for small value of $n \geq 1$ is not important.