Series Convergence Criterion

Question

The greatest value of p such that the series

\begin{equation} \sum_{n=1}^{\infty}(-1)^n \cdot \tan \left(\frac{1}{\sqrt{n^p}}\right) \cdot \ln \left(1+\frac{1}{n^{2 p}}\right) \end{equation}

converges conditionally, what is the value of p?

Answer 1

The sequence $$ \tan \left(\frac{1}{\sqrt{n^p}}\right) \cdot \ln \left(1+\frac{1}{n^{2 p}}\right)$$ is decreasing and convergent to $0$ for $p>0. $ Therefore the series is convergent due to the Leibniz test. The series is absolutely convergent for $p/2 + 2p>1$ as $$\tan \left(\frac{1}{\sqrt{n^p}}\right) \cdot \ln \left(1+\frac{1}{n^{2 p}}\right )\approx n^{-p/2-2p}$$

For $p<0$ the general term does not tend to $0,$ as the $\log $ factor tends to $+\infty$ and $\sin(n^{-p/2})$ is dense in $[-1,1].$ This a nontrivial fact discussed here

Answer 2

Here is the proof of the convergence of the series under consideration for p > 0.The simple-minded approach fails.

SumConvergence[(-1)^n*Tan[1/Sqrt[n^p]]*Log[1 + 1/n^(2*p)], n, 
Assumptions -> p > 0]

returns the input. In order to apply the alternating series test we find

Asymptotic[Tan[1/Sqrt[n^p]]*Log[1 + 1/n^(2*p)], n -> Infinity
 , Assumptions -> p > 0]

n^(-5 p/2)

Therefore, this tends to zero as n approaches infinity. Now we find

D[Tan[1/Sqrt[n^p]]*Log[1 + 1/n^(2*p)], n]

-((n^(-1 + p) p Log[1 + n^(-2 p)] Sec[1/Sqrt[n^p]]^2)/( 2 (n^p)^(3/2))) - (2 n^(-1 - 2 p) p Tan[1/Sqrt[n^p]])/(1 + n^(-2 p))

It's clear that both terms are negative for n>=1&&p>0 and

NMaximize[{-((n^(-1 + p) p Log[1 + n^(-2 p)] Sec[1/Sqrt[n^p]]^2)/(
2 (n^p)^(3/2))) - (2 n^(-1 - 2 p) p Tan[1/Sqrt[n^p]])/( 1 + n^(-2 p)),   n \[Element] PositiveIntegers && p > 0}, {n, p}

{-7.51567*10^-246, {n -> 28, p -> 67.9428}}

confirms it. Now we can conclude the convergence.

Answer 3

Applying Taylor's theorem to $\tan(x)$ and $\ln(1 + x)$ for $x$ around zero tells us that $$\tan(1/\sqrt{n^p})\ln(1 + 1/n^{2p}) \sim \frac{1}{n^{p/2 + 2p}} = n^{-5p/2}.$$ The limit comparison test applied to the series $\sum_{n=1}^\infty n^{-5p/2}$ tells us that the series $\sum_{n=1}^\infty (-1)^n \tan(1/\sqrt{n^p})\ln(1 + n^{-2p})$ is absolutely convergent for $p > 2/5$. Since $\tan(x)$ and $\ln(1+x)$ are monotonic functions, the alternating series test applied to this series implies that for $p > 0$, our series is conditionally convergent.

The answer is $p = 2/5$.

Answer 4

Hi all there is another way to solve this question

$$\underset{n\to \infty }{\mathop{\lim }}\,\frac{\tan \left( \frac{1}{\sqrt{{{n}^{p}}}} \right)\ln \left( 1+\frac{1}{{{n}^{2p}}} \right)}{\frac{1}{\sqrt{{{n}^{p}}}}\frac{1}{{{n}^{2p}}}}=1\\$$

Since the ratio 1 > 0, the two series have the same convergence properties.

$$\frac{1}{\sqrt{n^p} } \frac{1}{n^{2p}} =\frac{1}{n^{\frac{p}{2}+2p }} $$

Let

$$\frac{p}{2}+2p\le 1 \Rightarrow p\le\frac{2}{5} $$