Decide whether the following series converges $\sum_{n=1}^{\infty}\frac{(\ln n)^2}{n^{3/2}}$

Question

Looking for a neat and smart way to solve this.

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thankyou everyone, u guys deserve this http://i.imgur.com/sHxpw1t.png — loading...., Sep 01 '14 at 19:43
Possible duplicate of Convergence of $\sum\limits_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} $ for nonnegative $\alpha$ and $\beta$ — Nosrati, Nov 03 '18 at 06:58

score 3 · Answer 1 · answered Sep 01 '14 at 18:51

We have (we can prove it using the L'Hôpital's rule)

$$(\ln n)^2=o(n^{1/3})$$ so

$$\frac{(\ln n)^2}{n^{3/2}}=o\left(\frac1{n^{\frac32-\frac13}}\right)$$ and since $\frac32-\frac13>1$ we conclude by comparison with a Riemann convergent series that the given series is convergent.

Ma Ming · Answer 2 · 2014-09-01T19:50:11.737

2

Hint:

Claim 1. $$ \sum \frac{1}{n^\alpha}<\infty,\quad \forall \alpha>1 $$

Claim 2. For any $\beta >0$, $$ \ln n<n^\beta, \quad n\gg 1. $$

edited Sep 01 '14 at 19:50

answered Sep 01 '14 at 18:49

Ma Ming

7,522

score 2 · Answer 3 · answered Sep 01 '14 at 18:56

You know that there exist such a $N$, that for $n>N$ you have($\varepsilon>0$):

$$\ln n < n^{\varepsilon}$$

So for $n>N$:

$$\frac{(\ln n)^2}{n^{\frac{3}{2}}}<\frac{n^{2 \varepsilon}}{n^{\frac{3}{2}}}=\frac{1}{n^{\frac{3}{2}-2\varepsilon}}$$

If you choose $\varepsilon<\frac{1}{4}$, then $\frac{3}{2}-2\varepsilon>1$, so the series:

$$\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}-2\varepsilon}}$$

converges, so $\displaystyle \sum_{n=1}^{\infty}\dfrac{(\ln n)^2}{n^{3/2}}$ also converges.

score 0 · Answer 4 · edited Apr 13 '17 at 12:21

Here is an approach using Cauchy condensation theorem which allows us to consider instead the series

$$ \sum_{n=1}^{\infty} \frac{n^2}{2^{n/2}} $$

which is a convergent series if it is copared with the geometric series $\sum_{n}x^n$ with the condition of convergence $|x|<1$. See related techniques.

Cauchy Condensation Theorem:

For a positive non-increasing sequence $f(n)$, the sum $$ \sum_{n=1}^{\infty}f(n) $$ converges if and only if the sum $$\sum_{n=0}^{\infty} 2^{n}f(2^{n})$$ converges.

score 0 · Answer 5 · answered Sep 08 '14 at 08:09

well I was trying in this way,we know,$$\forall x\ge1 $$ $$\frac{1}{x}\le\frac{1}{x^\frac{4}{5}}$$so$$\int^n_1\frac{1}{x}dx\le\int^n_1\frac{1}{x^\frac{4}{5}}dx$$i.e.$$log(n)\le 5n^\frac{1}{5}-5<5n^\frac{1}{5}$$$\Rightarrow$$$(log(n))^2<25n^\frac{2}{5}$$$\Rightarrow$$$\frac{(log(n))^2}{n^\frac{3}{2}}<25\frac{1}{n^\frac{11}{10}}$$$\Rightarrow$$$\sum_{n=1}^{\infty}\frac{(log(n))^2}{n^\frac{3}{2}}<\sum_{n=1}^{\infty}25\frac{1}{n^\frac{11}{10}}$$ RHS converges and hence LHS converges $$\sum_{n=1}^{\infty}\frac{1}{n^t}$$converges iff t>1

Decide whether the following series converges $\sum_{n=1}^{\infty}\frac{(\ln n)^2}{n^{3/2}}$

5 Answers5