Convergence of the series using power series
Does the following series converge or diverge?
$$\sum_{n=1}^{\infty}\frac{{(-1)}^{n}}{n^{\frac2n}}$$
$$\sum_{n=1}^{\infty}\frac{{(-1)}^{n} (\ln n)^2}{n^{\frac12}}$$
I am trying to use the power series to do a direct comparison test and solve both of these questions. However the negative values in the numerator is throwing me off, i can't seem to find the right value to compare. Anyone could point me in the right direction?
sequences-and-series convergence
PhantomPhantom
$\begingroup$ "I have two questions" What are they? $\endgroup$ – Did Mar 27 '14 at 9:18
$\begingroup$ @Did The two questions which i have numbered as Qn1 and Qn2. "I have two questions" = "I am given two questions to solve". $\endgroup$ – Phantom Mar 27 '14 at 10:15
$\begingroup$ The things labelled Qn1 and Qn2 are not questions, but formulas of quantities depending on $n$. $\endgroup$ – Did Mar 27 '14 at 12:03
$\begingroup$ @Did Ah..okay, i get what you mean now. My mistake :) $\endgroup$ – Phantom Mar 27 '14 at 12:05
$\begingroup$ Still no question in Qn1 and Qn2. $\endgroup$ – Did Mar 27 '14 at 12:07
Notice that $$\lim_{n\to\infty}\frac{1}{n^{2/n}}=1\ne0$$ so the first series is divergent.
For the second series let $$g(x)=\frac{(\ln x)^2}{\sqrt x}$$ then using the derivative prove that this function is increasing (to $0$) for a sufficient large $x$ and conclude the convergence of this series using the alternating series test.
$\begingroup$ The first series does not converge. $\endgroup$ – 5xum Mar 27 '14 at 8:37
$\begingroup$ Nice work, Sami! $\endgroup$ – Namaste Mar 28 '14 at 11:44
Both series are so called alternating series. For them a very nice test shows if they converge. Here is the link to it.
For the first series, you may encounter some problems. Take a look at what the limit $$\lim_{n\to\infty}\frac{1}{n^{\frac2n}}$$ is.
5xum5xum
$\begingroup$ The first one is not alternating. Recall that the general term of an alternating series (1) goes to zero and (2) is decreasing in absolute value. $\endgroup$ – Did Mar 27 '14 at 12:04
$\begingroup$ @Did It is alternating. By definition (see en.wikipedia.org/wiki/Alternating_series), any series which has elements which switch their sign is alternating. What you described is the conditions that must be filled for an alternating series to be convergent. $\endgroup$ – 5xum Mar 27 '14 at 13:16
$$\lim_{n\to\infty}{n^{\frac 2n}}=1,$$ so $$\frac{{(-1)}^{n}}{n^{\frac2n}}\not\to 0$$ and the first series does not converge.
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
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