# 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.

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!

$$\boldsymbol{\frac{n^{1/n}}{\log(n)}}$$ 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 $$\newcommand{\li}{\operatorname{li}} \large\scriptstyle g_1(n)=\li(n)+\log(n)+\frac1{2\log(n)}-\frac{\log(n)}{2n}-\frac1{12n\log(n)^2}-\frac{\log(n)^2}{12n^2}+\frac{\log(n)}{6n^2}-\frac1{8n^2}+\cdots\tag8$$ 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 $$\sum_{k=2}^\infty(-1)^k\frac{k^{1/k}}{\log(k)}=1.287832248273636925802210499651\tag{12}$$

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