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What test do you use to prove that

$$\sum_{n=2}^\infty \frac{\ln(n)}{n^{3/2}}$$

converges?

I tried the limit comparison test using $\frac{1}{n^{3/2}}$ as the comparison, but it did not converge.

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  • $\begingroup$ But $\sum\limits_{n=2}^\infty \frac{1}{n^{3/2}}$ does converge.... $\endgroup$
    – The Count
    Jan 10, 2017 at 0:16

3 Answers 3

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Apply Cauchy Condensation Test.

Let $a_n = \dfrac{\ln(n)}{n^{3/2}}$, then $a_{2^n} = \dfrac{\ln 2^n }{(2^n)^{3/2}}$

Therefore, $$a_{2^n} = \dfrac{\ln 2^n }{(2^n)^{3/2}} = \frac{n\ln 2 }{2^{3n/2}} \leq \ln2\frac{n}{2^n}$$

Thus $$ \sum_{n=2}^{\infty}\dfrac{\ln 2^n }{(2^n)^{3/2}} \leq \sum_{n=2}^{\infty} \ln2\frac{n}{2^n}.$$

The RHS is a convergent series, by the Ratio Test, as you should verify.

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  • $\begingroup$ Neat footwork there :-) $\endgroup$ Apr 30, 2014 at 13:17
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Hint: You can use integral test.

Added: You need to consider the integral

$$ \int_{2}^{\infty} \frac{\ln(x)}{x^{3/2}}dx , $$

which can be integrated using integration by parts.

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  • $\begingroup$ Using the integral test I got the answer to be = 25.55275762. $\endgroup$
    – khap93
    Apr 30, 2014 at 2:49
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    $\begingroup$ @khap93, you are not supposed to directly calculate the sum...well I guess that shows you the sum does converge. $\endgroup$
    – IAmNoOne
    Apr 30, 2014 at 2:49
  • $\begingroup$ @Nameless Once you take the integral how are you supposed to prove it is convergent? $\endgroup$
    – khap93
    Apr 30, 2014 at 2:53
  • $\begingroup$ @khap93, well if you were able to take the integral and get something finite, this means the integral converges, and therefore the original sum most also converge. $\endgroup$
    – IAmNoOne
    Apr 30, 2014 at 2:55
  • $\begingroup$ @khap93: Read the main result well that I referred you to. Do not rush. $\endgroup$ Apr 30, 2014 at 2:57
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You can use that $\ln n$ goes to infinitely much more slowly that any (positive) power of $n$, that is: for every $\alpha>0$, $\ln n < n^\alpha$ for $n$ large enough. You can apply this to $\alpha=1/4$ says (any $\alpha$ strictly between $0$ and $1/2$ will do, so let's say $1/4$), so $\ln n < n^{1/4}$ for $n$ large enough, hence $\ln n / n^{3/2} < 1/n^{5/4}$ for $n$ large enough, and since the series $\sum 1/n^{5/4}$ converges (as does $\sum 1/n^\beta$ for any $\beta>1$), you're done.

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