I know that the series $\sum_{n=2}^{\infty} \dfrac{{log(n^b)}^{M}}{n^b}$, where 1 < b < $\infty$ and M is an arbitrary nonnegative integer converges, but I don't know what I should compare the series to. Thanks for any help in advance.
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there is a special notation for that : $\log(x) = \mathcal{O}(x^\epsilon)$ so $(\log(n^b))^M = \mathcal{O}(n^{\epsilon})^ M = \mathcal{O}(n^\epsilon)$ – reuns Oct 12 '16 at 04:10
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How does that help? Could you put it in the form of an answer? – shmiggens Oct 12 '16 at 04:14
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for every $\epsilon > 0$, $\log(x) = \mathcal{O}(x^\epsilon)$, that is for every $\epsilon > 0$, there exists a $C$ such that $\log(x) < C x^\epsilon$ for every $x > 1$ (that's what Dr.MV wrote) – reuns Oct 12 '16 at 04:17
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Could you put this in the form of an explicit answer? I'm not seeing it. – shmiggens Oct 12 '16 at 04:20
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of course I won't write a course on the big-O notation just for you – reuns Oct 12 '16 at 04:20
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@user1952009 I provided a solution roughly 40 minutes ago that developed what is tantamount to your concise approach. The OP might be confused between the equivalence (perhaps notationally?). – Mark Viola Oct 12 '16 at 04:38
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In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequality
$$\bbox[5px,border:2px solid #C0A000]{\log(x)\le x-1<x} \tag 1$$
for $x>0$.
Using $(1)$ along with the property that $\log(x^a)=a\log(x)$, we find that for $a\ge 0$
$$\log(x)\le \frac{x^a}{a}$$
For any $M\ge0$, choose $a$ such that $aM-b<-1$ (or, $a<\frac{b-1}{M}$). Then, we have
$$ \sum_{n=2}^N\frac{\log^M(n^b)}{n^b}\le \left(\frac ba\right)^M \sum_{n=2}^Nn^{aM-b} \tag 2$$
By the comparison test (or integral test), the sum on the right-hand side of $(2)$ converges and hence by comparison the integral on the left-hand side converges also.
Mark Viola
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The series $\sum_{n=2}^\infty \frac{1}{n^{1+\epsilon}}$ converges for all $\epsilon>0$. Now note that $b-aM=1+(\epsilon)$ where $\epsilon = b-aM-1>0$. – Mark Viola Oct 12 '16 at 04:19
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I like your solution, but I'm not seeing where a few things are coming from. For example, where is the fraction (b/a)^M coming from? – shmiggens Oct 12 '16 at 04:47
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$\log^M(n^b)=b^M\log^M(n)$ and $\log^M(x)\le \left(\frac{x^a}{a}\right)^M=\frac{x^{aM}}{a^M}=\frac{1}{a^M}x^{aM}$ – Mark Viola Oct 12 '16 at 04:57
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I should mention that this series was part of a larger problem that involved showing that the list of a sequence of random variables is equal to infinity almost certainly. – shmiggens Oct 12 '16 at 15:55
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I personally think the more efficient way to show the convergence of the series is via Cauchy condensation test which states:
(Cauchy's Condensation Test) For a non-negative non-increasing sequence $\{a_n\}$, the series $\sum a_n$ converges iff $\sum 2^k a_{2^k}$ converges.
Observe \begin{align} \sum^\infty_{k=1} 2^k \frac{(\log 2^{bk})^M}{2^{bk}} = \sum^\infty_{k=1}\frac{(bk \log 2)^M}{2^{(b-1)k}}=(b\log 2)^M\sum^\infty_{k=1}\frac{k^M }{2^{(b-1)k}} <\infty \end{align} which converges iff $b>1$.
Jacky Chong
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"more efficient" than what? And how did you prove that the last series converges? What test did you use then?? – Mark Viola Oct 12 '16 at 04:35
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@Dr.MV I personally think the method is more efficient than comparison test when dealing with series that have logarithms. However, it's not in general better than comparison test. – Jacky Chong Oct 12 '16 at 04:38
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See my solution. It is concise, efficient, and doesn't rely on appeal to the CCT. Moreover, you don't state how you know that your final series converges. So, what test did you use there? – Mark Viola Oct 12 '16 at 04:40
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@Dr.MV Don't get me wrong. I do like your solution but it's my personal taste that CCT is in general a nice test to use. Using CCT, I could easily show when the harmonic series converges which is one of the reasons why I like CCT. – Jacky Chong Oct 12 '16 at 04:47