Question

Convergence or divergence

Answer 1

\[\sum_{n=2}^{\infty} \frac{ 1 }{ (\ln n)^{\ln n} }\]

Answer 2

What should I use for the comparison test here, an and bn, I'm not sure.

Answer 3

Well bn

Answer 4

\[a_n = \frac{ 1 }{ (\ln n)^{\ln n} }\]

Answer 5

?

Answer 6

[ln(n)]^(ln(1/n))

it seems to be O(x^2) but i got no idea if thats helpful or not

Answer 7

I suspect it's converging so should I just try 1/n^2 or something?

Answer 8

1/y <= 1/n^2

n^2 <= y assuming y is positive

Answer 9

Looks good I think

Answer 10

\[\frac{ 1 }{ n^2 } > \frac{ 1 }{ (\ln n)^{\ln n} } \implies (\ln n)^{\ln n} > n^2\]
\[\ln(\ln n)^{\ln n} > \ln n^2 \implies n> e^{e^{2}}\]

It would be convergent?

Answer 11

ln(x) ^ ln(x) is not always < n^2

Answer 12

if p >1
actually I'll just use ratio or root test or something comparison tests as such always bother me as I don't fully understand them haha.

Answer 13

e^(ax) <= ln(x) ^ ln(x)

ax <= ln^2(x)

Answer 14

does e^(-n) converge? it seems to be bigger than ln(x) ^(-ln(x)) after say x=.5

Answer 15

I don't think so

Answer 16

this is breaking the wolf ...

Answer 17

Haha xD, I don't think ratio test is working either, I'm so confused.

Answer 18

whats wrong with comparison test with 1/n^2 ?

Answer 19

Was it right haha? I was guessing there.

Answer 20

After skipping few initial terms,
each term in 1/n^2 series is larger than the corresponding term in 1/(ln(n))^(ln(n))

yes ?

Answer 21

Yup

Answer 22

since the series with larger terms 1/n^2 converges by p-series test,
the terms with smaller terms 1/(ln(n))^(ln(n)) also converges

Answer 23

that makes sense intuitively atleast right ?

Answer 24

Note that the initial terms dont matter for convergence

Answer 25

http://www.wolframalpha.com/input/?i=x%5E2+%3D+ln%28x%29%5E%28ln%28x%29%29

yeah, i was mixing the 2. the bigger value in fraction form is the smaller number

Answer 26

can we prove its cauchy?

Answer 27

i read somewhere that its easier to prove cauchy then convergence, and that every cauchy is convergent

Answer 28

But ln(ln n) is an increasing function right, wouldn't that be diverging, making the whole series diverge?

Answer 29

why are we talking about ln(ln(n)) ?

Answer 30

\[\ln(\ln n) = 2 \implies \ln n = e^2 \implies n = e^{e^{2}}\]

and \[n = e^{e^{2}}\] is some huge number as well..wait nvm the larger series is converging.

Answer 31

above arguement with 1/n^2 is sufficient,
we don't need to explicitly find a "n" value for proving the convergence

Answer 32

So is there also other ways of working this problem out without it failing?

Answer 33

Yeah I got that

Answer 34

it doesnt matter how large the n have to be for 1/n^2 terms to dominate...
it is fine as long as there exists some finite n, after which all the terms in 1/n^2 are larger than the terms in the particular series in question..

Answer 35

i think we're trying comparison after giving up on all other tests eh ?

Answer 36

Lool, yeah, I tried ratio only other than this and it failed, but I think root might work, I'm just messing around with this question trying to understand what exactly is going on. But, what you said makes a lot of sense. The 1/n^2 is always dominating and the larger series converges because it's a p series with p = 2>1.

Answer 37

most importantly - comparison test works only when the terms in both series are positive

Answer 38

Right!

Answer 39

Thank you @amistre64 and @ganeshie8