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I was wondering if someone knows a comparison test argument for the convergence of $$\sum_{i=2}^{\infty} \frac{(\log(i))^k}{i^2}$$ for all $k \gt 0$. If possible, I am particularly interested in a comparison test argument. When I put the series into Wolfram Alpha, it told me that the series converges by the comparison test, however, I cannot come up with a nice series we can compare it with.

Thanks! Lauren

(Note: $\log$ is the natural logarithm in this case)

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  • $\begingroup$ Can you show that if $a_n = \frac{(\ln n)^k}{n^2}$ and $b_n = \frac{1}{n^{3/2}}$, then $\frac{a_n}{b_n} \xrightarrow[n\to\infty]{} 0$? $\endgroup$ – Clement C. Oct 21 '15 at 23:15
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No matter how small you take $\delta > 0,$ you always have $$ \lim_{n \rightarrow \infty} \frac{\log n}{n^\delta} = 0. $$ For your $k,$ take $\delta = \frac{1}{2k}$ for example.

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  • $\begingroup$ Thanks for your response. I can see this intuitively, but, is there any way to show this a little more formally? $\endgroup$ – Lauren Hayes Oct 21 '15 at 23:18
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    $\begingroup$ lhospitals rule $\endgroup$ – Will Jagy Oct 21 '15 at 23:18
  • $\begingroup$ Apologies for the ambiguous question, I meant for all $k \gt 0$ $\endgroup$ – Lauren Hayes Oct 21 '15 at 23:19
  • $\begingroup$ @LaurenHayes If you know how to prove it for $k=1$, you know it for any $k$. Indeed, just take the $k$-th root, and replace $\delta$ by $\delta/k$. $\endgroup$ – Clement C. Oct 21 '15 at 23:29
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Hint One can use that $\log x$ is $O(x^{\epsilon})$, and hence $(\log x)^k$ is $O(x^{k \epsilon})$ for all $\epsilon > 0$. Thus, the summand is $O(x^{k \epsilon - 2})$, and we can choose $\epsilon$ small enough so that the sum converges.

Alternatively, we can apply the integral test; the substitution $x = \exp u$ transforms the integral to a convenient form and, using a common integral definition of the Gamma function, should give a reasonably good upper bound in terms of something like $\Gamma(k)$.

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There have been several solid answers that rely on the fact that for all $\alpha >0$, $\log x=O(x^{\alpha})$. I thought it would be instructive to show this using standard tools that don't rely on L'Hospital's Rule but rather, only involve the integral definition of the log and simple inequalities.

To that end, the (integral representation) definition of the log function can be written as

$$\log x\equiv\int_1^x\frac{1}{t}\,dt$$

Therefore, for any number $\alpha>0$, however small, we have

$$\log x^{\alpha}=\int_1^{x^{\alpha}}\frac{1}{t}\,dt \tag 1$$

Now, we can easily find an upper bound for $\log x$ using $(1)$. To that end, we write

$$\begin{align} \alpha \log x&=\log x^{\alpha}\\\\ &=\int_1^{x^{\alpha}}\frac{1}{t}\,dt\\\\ &\le x^{\alpha}-1\\\\ &<x^{\alpha} \end{align}$$

whereupon dividing $(2)$ by $\alpha$ yields

$$\log x\le \frac{x^{\alpha}}{\alpha} \tag 3$$

Now, recall that $(3)$ holds for all $\alpha >0$. Then, given a number $k$, we can choose this $\alpha$ so that $\alpha k=\delta <1$. Therefore, exploiting $(3)$ we have

$$\frac{\log^k i}{i^2}\le\frac{i^{\alpha k}}{\alpha i^2}=\frac{1}{\alpha i^{2-\delta}}$$

and therefore the series $\sum_{i=1}^\infty\frac{\log^k i}{i^2}$ converges by the comparison test since $2-\delta >1$. And we are done.

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. - Mark $\endgroup$ – Mark Viola Nov 22 '15 at 6:24

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