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

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?

• Did you mean $l_n$ and what is it? – Will M. Nov 24 '18 at 1:11
• If you meant $\log(n),$ write that \log(n). – Will M. Nov 24 '18 at 1:12
• How did you show that the positive series does not converge? (This informs how we should address your main question.) – Eric Towers Nov 24 '18 at 1:20
• Note that $n^{1/n}=1+\frac{\log(n)}n+O\!\left(\frac{\log(n)^2}{n^2}\right)$ – robjohn Nov 24 '18 at 2:54

\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.

$$\boldsymbol{\frac{n^{1/n}}{\log(n)}}$$ 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 $$n\ge2$$.

$$\boldsymbol{\frac{n^{1/n}}{\log(n)}}$$ 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 $$\lim_{n\to\infty}\frac{n^{1/n}}{\log(n)}=0\tag5$$

Convergence

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

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 $$\newcommand{\li}{\operatorname{li}} \largeg_1(n)=\li(n)+\log(n)+\frac1{2\log(n)}-\frac{\log(n)}{2n}-\frac1{12n\log(n)^2}-\frac{\log(n)^2}{12n^2}+\frac{\log(n)}{6n^2}-\frac1{8n^2}+\cdots\tag8$$ Applying Euler-Maclaurin to $$f(2n)$$ yields $$\hspace{-4pt}\largeg_2(n)=\frac12\li(2n)+\frac12\log(2n)+\frac1{2\log(2n)}-\frac{\log(2n)}{8n}+\frac1{8n}-\frac1{12n\log(2n)^2}-\frac{\log(2n)^2}{96n^2}+\frac{5\log(2n)}{96n^2}-\frac3{64n^2}+\cdots\tag9$$ 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 $$\sum_{k=2}^\infty(-1)^k\frac{k^{1/k}}{\log(k)}=1.287832248273636925802210499651\tag{12}$$

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.

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

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