Find the range of p for which the series $\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^p}$ is convergent. I want to use dyadic test rather than integral test. Anyone can help?
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Anyone know how to make the LaTex code work? This is my first time using it here. – Steve May 13 '14 at 03:25
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Put $$ around it. – Daniel Donnelly May 13 '14 at 03:27
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I edited post, please check formula – Daniel Donnelly May 13 '14 at 03:28
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What's dyadic test, I can't google & find it. – Daniel Donnelly May 13 '14 at 03:29
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Thank you! Dyadic Test: Suppose $(a_n)$ is a non-increasing sequence and $a_n \ge 0$, then $\sum a_k$ converges if and only if $\sum 2^k\times a_{2^k}$ converges. – Steve May 13 '14 at 03:37
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What Steve calls "dyadic test" is what is also known, perhaps more commonly, as (Cauchy's) Condensation Test. – DonAntonio May 13 '14 at 03:39
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Possible duplicate of Convergence of the series $\sum \limits_{n=2}^{\infty} \frac{1}{n\log^s n}$ – Arnaud D. Dec 05 '19 at 10:42
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By the Cauchy Condensation Test, which is probably what you mean by a dyadic test, a decreasing series of positive terms $\sum_{n=2}^\infty f(n)$ converges if and only if $\sum_{k=2}^\infty 2^k f(2^k)$ converges.
In our case, $f(n)=\frac{1}{n(\ln n)^p}$, and therefore
$$\sum_{k=2}^\infty 2^k f(2^k)=\sum_{k=2}^\infty \frac{1}{(\ln 2)^pk^p}.$$
Now we can use the fact that $\sum_2^\infty \frac{1}{k^p}$ converges if and only if $p\gt 1$.
Remark: If you want to prove that $\sum_{k=2}^\infty \frac{1}{k^p}$ converges iff $p\gt 1$, you can use Cauchy Condensation once more. One ends up with a geometric series.
André Nicolas
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