4

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.

Martin Sleziak
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khap93
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3 Answers3

7

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.

qqo
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6

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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Mhenni Benghorbal
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  • Using the integral test I got the answer to be = 25.55275762. – khap93 Apr 30 '14 at 02:49
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    @khap93, you are not supposed to directly calculate the sum...well I guess that shows you the sum does converge. – qqo Apr 30 '14 at 02:49
  • @Nameless Once you take the integral how are you supposed to prove it is convergent? – khap93 Apr 30 '14 at 02:53
  • @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. – qqo Apr 30 '14 at 02:55
  • @khap93: Read the main result well that I referred you to. Do not rush. – Mhenni Benghorbal Apr 30 '14 at 02:57
  • @Nameless Thank you both. I was making the mistake when using integration by parts of forgetting that x^(3/2) was on the bottom and this = x(-3/2) which threw off my integration. – khap93 Apr 30 '14 at 03:04
6

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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