What is the limit value of $a$ that $\sum\limits_{k = 1 }^ \infty \frac{ \ln(k)}{k^a}$ converges

if a>1 then the series $\sum\limits_{k = 1 }^ \infty \frac{ 1}{k^a}$ converges

What is the limit value of $a$ for the series below?

$$\sum\limits_{k = 1 }^ \infty \frac{ \ln(k)}{k^a}$$

My steps to find the value: if $x > 1$ then $\ln x <x$

thus

$$\sum\limits_{k = 1 }^ \infty \frac{ \ln(k)}{k^a} < \sum\limits_{k = 1 }^ \infty \frac{ k}{k^a} = \sum\limits_{k = 1 }^ \infty \frac{ 1 }{k^{a-1}}$$

if $a>2$, It is sure that $\sum\limits_{k = 1 }^ \infty \frac{ \ln(k)}{k^a}$ the series converges. But we can find a lower value for the series.

Could you please help me to find the exact value of $a$ that the series converge?

It can easily proved that for all $\varepsilon>0$ there exists $x_0$ such that for all $x>x_0$: $$\ln x <x^\varepsilon$$

It can be shown in couple of ways (i.e calculating $\lim_{x\to \infty}\frac{\ln x}{x^\varepsilon}$)*.

Now, for any given $a>1$ we can write $a=1+2\delta$ and from the previous statement we can find a natural $N$ such that for every $n>N$: $$\ln n<n^\delta$$

So we get: $$\sum_{n=N+1}^\infty \frac{\ln n}{n^a} = \sum_{n=N+1}^\infty \frac{\ln n}{n^{1+2\delta}}<\sum_{n=N+1}^\infty \frac{ n^\delta}{n^{1+2\delta}}=\sum_{n=N+1}^\infty \frac{1}{n^{1+\delta}}$$ The latter sum is a "p-series" tail with p>1 and therefor converges.

*Addition - $$\lim_{x\to \infty}\frac{\ln x}{x^\varepsilon}\overset{\text{L'Hopital}}{=\!=\!=\!=\!=\!=}\lim_{x\to \infty}\frac{\frac{1}{x}}{\varepsilon \cdot x^{\varepsilon-1}}=\frac {1}{\varepsilon}\lim_{x\to \infty}\frac{1}{x^\varepsilon}=0$$

• Can you figure out $\lim_{x\to \infty}\frac{\ln x}{x^\varepsilon}$? – Amihai Zivan Jul 25 '12 at 14:02
• need LHospital rule to proof. Right? – Mathlover Jul 25 '12 at 14:04
• yup - that's one of the ways doing it. – Amihai Zivan Jul 25 '12 at 14:10

Let $f(x)=e^x-1-x$, then $f'(x)=e^x-1$ and the only $x\in\mathbb{R}$ so that $e^x-1=0$ is $x=0$. Note that $f''(0)=1$ so $f(x)$ reaches a minimum of $0$ at $x=0$. Therefore, for all $x\in\mathbb{R}$, $$1+x\le e^x\tag{1}$$

Thus, we get that for all $x\in\mathbb{R}$, $x<e^x$, and therefore, for all $x>0$, $$\log(x)< x\tag{2}$$ For any $\alpha>0$, apply inequality $(2)$ to $x^\alpha$ and divide by $\alpha$ to get $$\log(x)< \tfrac1\alpha x^\alpha\tag{3}$$ Then for any $\epsilon>0$, we have \begin{align} \sum_{k=1}^\infty\frac{\log(k)}{k^{1+\epsilon}} &<\sum_{k=1}^\infty\frac{\frac2\epsilon k^{\epsilon/2}}{k^{1+\epsilon}}\\ &=\frac2\epsilon\sum_{k=1}^\infty\frac1{k^{1+\epsilon/2}}\tag{4} \end{align} and $(4)$ converges for all $\epsilon>0$. Thus, for any $\alpha>1$, $$\sum_{k=1}^\infty\frac{\log(k)}{k^\alpha}\tag{5}$$ converges.

Alternatively, since $\frac{\log(x)}{x^\alpha}$ is a decreasing function when $x\ge e$ and $\alpha\ge1$, \begin{align} \sum_{k=3}^\infty\frac{\log(k)}{k^\alpha} &\le\int_1^\infty\frac{\log(x)}{x^\alpha}\mathrm{d}x\\ &=\int_0^\infty\frac{x}{e^{\alpha x}}\,e^x\,\mathrm{d}x\\ &=\frac1{(\alpha-1)^2}\int_0^\infty x\,e^{-x}\,\mathrm{d}x\\ &=\frac1{(\alpha-1)^2}\tag{6} \end{align} Thus, for any $\alpha>1$, we again have that $(5)$ converges.

It converges for $a\gt 1$. Try the integral test.

Here's another approach. Let $\displaystyle c_k = \frac{\log k}{k^a}$. Examine the ratio of successive terms for large $k$, $$\frac{c_{k+1}}{c_{k}} = 1 - \frac{a}{k} + \frac{1}{k\log k} + O\left(\frac{1}{k^2}\right).$$ By Raabe's test the series converges for $a>1$ and diverges for $a<1$. By Bertrand's test the series diverges for $a=1$.

Raabe's test tells us that if $$\left|\frac{a_{n+1}}{a_{n}}\right| \sim 1 - \frac{s}{n} \hspace{5ex}(n\to\infty),$$ then the series converges absolutely if $s>1$, diverges if $s<1$, and may converge or diverge if $s=1$.

A simplified version of Bertrand's test tell us that if $$\left|\frac{a_{n+1}}{a_{n}}\right| \sim 1 - \frac{1}{n} - \frac{s}{n\log n} \hspace{5ex}(n\to\infty),$$ then the series converges absolutely if $s>1$, diverges if $s<1$, and may converge or diverge if $s=1$.

All you need to have for your serie to converge is $a>1$. Indeed, $\displaystyle\frac{\ln k}{k^a} = \frac{\ln k}{k^{\frac{a-1}{2}}}\times\frac{1}{k^{1+\frac{a-1}{2}}} = o(\frac{1}{k^{1+\frac{a-1}{2}}})$.

Case 1: $a>1$. Pick $b \in (1,a)$.

Then

$$\lim_k \frac{\frac{ \ln(k)}{k^a}}{\frac{1}{k^b}}=0 \,.$$

Since $\sum_k \frac{1}{k^b}$ is convergent, by the Limit Comparison Theorem your series is convergent.

Case 2: $a \leq 1$.

Then

$$\lim_k \frac{\frac{ \ln(k)}{k^a}}{\frac{1}{k^a}}=\infty \,.$$

Since $\sum_k \frac{1}{k^a}$ is divergent, then by the Limit Comparison Theorem your series is divergent.