Convergence of the series $\sum_{n=1}^{\infty}\frac{(-1)^n\sqrt[n]n}{\log\ \ n}$.

Question

I am analizing the convergence, absolute convergence and conditional convergence of the series $\sum_{n=1}^{\infty}\frac{(-1)^n\sqrt[n]n}{\log\ \ n}$. I proved already that the series $\sum_{n=1}^{\infty}\frac{\sqrt[n]n}{\log\ \ n}$ does not converges but I have not been able to prove if the original series converges or does not. Could someone help me with this, please?

Answer 1

\begin{align} \frac{d}{dx} \left( \frac{x^\frac1x}{\ln x}\right) &= \frac{x^{\frac1x-2}(1-\ln x)\ln x - x^{\frac1x-1}}{(\ln x)^2} \\ &=- \frac{x^{\frac1x-2}(x+(\ln x)^2 - \ln x)}{(\ln x)^2} <0 \end{align}

for $x > 1$.

Also, $$\lim_{n \to \infty} \frac{\sqrt[n]{n}}{\log n}=0$$

Hence, by alternating series test, it converges.

Remark: as pointed out by the rest, you might like to consider the series starting from the second term.

Answer 2

Hint: have a look at the alternating series test...(en.m.wikipedia.org/wiki/Alternating_series_test).

Answer 3

The sum does not exist because of the singularity at more than one evaluation points

You can do the comparison test and see that the series diverges.

Answer 4

$\log(1) = 0$. Hence you can't even do the first term!

Answer 5

$\boldsymbol{\frac{n^{1/n}}{\log(n)}}$ is decreasing for $\boldsymbol{n\ge2}$

Bernoulli's Inequality says $$ \begin{align} \frac{(n+1)^n}{n^{n+1}} &=\frac{\left(1+\frac1n\right)^{n+1}}{n+1}\\ &=\frac1{n+1}\left[\color{#C00}{\left(1-\frac1{n+1}\right)^{\frac{n+1}2}}\right]^{-2}\\ &\le\frac1{n+1}\left[\color{#C00}{\frac12}\right]^{-2}\\[9pt] &=\frac4{n+1}\tag1 \end{align} $$ Therefore, for $n\ge3$, $$ \begin{align} \frac{(n+1)^{\frac1{n+1}}}{n^{\frac1n}} &\le\left(\frac4{n+1}\right)^{\frac1{n(n+1)}}\\ &\le1\tag2 \end{align} $$ Thus, for $n\ge3$, $n^{1/n}$ is a decreasing function. Therefore, by verifying that $\frac{3^{1/3}}{\log(3)}\lt\frac{2^{1/2}}{\log(2)}$, we have $\frac{n^{1/n}}{\log(n)}$ is a decreasing function for $n\ge2$.

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$\boldsymbol{\frac{n^{1/n}}{\log(n)}}$ tends to $\boldsymbol{0}$

Bernoulli's Inequality says that $(1+\epsilon)^n\ge1+n\epsilon$. Thus, for any $\epsilon\gt0$, $$ \begin{align} \lim_{n\to\infty}n^{1/n} &\le\lim_{n\to\infty}\frac{1+\epsilon}{\epsilon^{1/n}}\\ &=1+\epsilon\tag3 \end{align} $$ Therefore, $$ \lim_{n\to\infty}n^{1/n}=1\tag4 $$ which implies $$ \lim_{n\to\infty}\frac{n^{1/n}}{\log(n)}=0\tag5 $$

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Convergence

Since $\frac{n^{1/n}}{\log(n)}$ monotonically tends to $0$, Dirichlet's Test guarantees the convergence of $$ \sum_{n=2}^\infty(-1)^n\frac{n^{1/n}}{\log(n)}\tag6 $$

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Evaluation

Application of the Euler-Maclaurin Sum Formula is not as straightforward as one might hope for this series. First, we define the function $$ \begin{align} f(n) &=\frac{n^{1/n}}{\log(n)}\\ &\sim\frac1{\log(n)}\left(1+\frac{\log(n)}n+\frac{\log(n)^2}{2n^2}+\frac{\log(n)^3}{6n^3}+\cdots\right)\tag7 \end{align} $$ Applying Euler-Maclaurin to $f(n)$ yields $$ \newcommand{\li}{\operatorname{li}} \large\scriptstyle g_1(n)=\li(n)+\log(n)+\frac1{2\log(n)}-\frac{\log(n)}{2n}-\frac1{12n\log(n)^2}-\frac{\log(n)^2}{12n^2}+\frac{\log(n)}{6n^2}-\frac1{8n^2}+\cdots\tag8 $$ Applying Euler-Maclaurin to $f(2n)$ yields $$ \hspace{-4pt}\large\scriptstyle g_2(n)=\frac12\li(2n)+\frac12\log(2n)+\frac1{2\log(2n)}-\frac{\log(2n)}{8n}+\frac1{8n}-\frac1{12n\log(2n)^2}-\frac{\log(2n)^2}{96n^2}+\frac{5\log(2n)}{96n^2}-\frac3{64n^2}+\cdots\tag9 $$ To get the even terms minus the odd terms we compute $$ \begin{align} \sum_{k=2}^{2n}(-1)^k\frac{k^{1/k}}{\log(k)} &=2g_2(n)-g_1(2n)+C\\ &=C+\frac1{2\log(2n)}+\frac1{4n}-\frac1{8n\log(2n)^2}+\cdots\tag{10} \end{align} $$ The constant $C$ is required since the Euler-Maclaurin sum formula has a constant of summation for much the same reason as integration does; it allows us to adjust for the beginning of the summation.

Looking at $(10)$, it is apparent that we have $$ \begin{align} C &=\sum_{k=2}^\infty(-1)^k\frac{k^{1/k}}{\log(k)}\\ &\sim\sum_{k=2}^{2n}(-1)^k\frac{k^{1/k}}{\log(k)}-2g_2(n)+g_1(2n)\tag{11} \end{align} $$ Extending $(7)$, $(8)$, and $(9)$ to include all terms with $n^{10}$ in the denominator and using $n=1000$ in $(11)$ yields $$ \sum_{k=2}^\infty(-1)^k\frac{k^{1/k}}{\log(k)}=1.287832248273636925802210499651\tag{12} $$