Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Thanks for answers

share|improve this question

6 Answers 6

up vote 4 down vote accepted

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

share|improve this answer
    
Thanks for answer. Could you please add the prove of your first sentence in your answer? –  Mathlover Jul 25 '12 at 12:30
    
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
    
Could you please add it to your answer? It would be perfect to see in your answer. Thanks a lot for your help. –  Mathlover Jul 25 '12 at 14:12

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.

share|improve this answer

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

share|improve this answer

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 $s=1$.

share|improve this answer

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.