1
$\begingroup$

I'm having difficulties examining convergence of the series

$$\sum_{n=2}^\infty \frac{1}{\sqrt{n\log n}}.$$

This root/ratio test is out of the question here. I tried to manipulate the denominator using properties of the logarithm but it didn't help.

Note: I'm not allowed to use the integral test (I also think that integration would be quiet challenging here).

$\endgroup$
  • 1
    $\begingroup$ Try the comparison test. $\endgroup$ – Joe Sep 10 '16 at 21:22
3
$\begingroup$

Since $\log n <\sqrt{n}$ for $n\geq2$, you have $$\frac{1}{\sqrt{n\log n}} > \frac{1}{\sqrt{n}\sqrt{n}} = \frac{1}{n}.$$ This sum of the last guy diverges.

$\endgroup$
1
$\begingroup$

note that $\log { n } <\sqrt { n } ,n\ge 2$ so $$\sum _{ n=2 }^{ \infty } \frac { 1 }{ \sqrt { n\log n } } >\sum _{ n=2 }^{ \infty } \frac { 1 }{ n } $$ means series diverges

$\endgroup$
1
$\begingroup$

Observe that $$ \sum_{i=2}^\infty\frac{1}{n^2}\le \sum_{i=2}^\infty\frac{1}{n\log n}\implies \sum_{i=2}^\infty\frac{1}{n}\le \sum_{i=2}^\infty\frac{1}{\sqrt{n\log n}} $$ Since, the harmonic series diverges which means the given series diverges.

$\endgroup$
1
$\begingroup$

$lim_n{{log(n)}\over {n^{1/4}}}={1\over 4}lim_n{{log(n^{1/4})}\over n^{1/4}}={1\over 4}lim_n {{log(n)}\over n}=0$, there exists $N$ such that$n>N$ implies that $long(n)<n^{1/4}$, we deduce that ${1\over\sqrt{nlog(n)}}>{1\over{n^{3/4}}}$. Compare with the sequence $\sum_n{1\over n^{3/4}}$.

$\endgroup$
1
$\begingroup$

You can use the cauchy condensation test which works great with logs. Let $a_n$ be the general term of your series. Then your series converges iff the series $\sum 2^n{a_{2^n}}$ converges. But $$ 2^n{a_{2^n}}=\frac{2^n}{\sqrt{2^n\log(2^n)}}=\frac{2^{n/2}}{\sqrt n\sqrt{\log2}}\longrightarrow \infty$$ so that your series diverges.

$\endgroup$

Your Answer

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