4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

7
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Neat footwork there :-) $\endgroup$ Apr 30, 2014 at 13:17
6
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Using the integral test I got the answer to be = 25.55275762. $\endgroup$
    – khap93
    Apr 30, 2014 at 2:49
  • 1
    $\begingroup$ @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
6
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.