Question

does the series diverge or converge?

Answer 1

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

Answer 2

@rational could you help me with this one?

Answer 3

could comparison work for this one?

Answer 4

could you show me how to do it?

Answer 5

i would bring down the ln exponent

Answer 6

This might help.
$$
\Large
{\sum_{n=2}^{\infty} \frac{ 1 }{ \ln n ^{\ln n} }
\\=\sum_{n=2}^{\infty} \frac{ 1 }{ \ln n \cdot \ln n }
}
$$

Answer 7

so you multiply the two?

Answer 8

\[\frac{ 1 }{ \ln n^2 }\]

Answer 9

thats incorrect

Answer 10

$$
\Large
{\sum_{n=2}^{\infty} \frac{ 1 }{ \ln n ^{\ln n} }
\\=\sum_{n=2}^{\infty} \frac{ 1 }{ \ln n \cdot \ln n }
\\=\sum_{n=2}^{\infty} \frac{ 1 }{ (\ln n )^2}
}

$$

Answer 11

and then compared that to 1/n?

Answer 12

we can use this argument
$$ \Large \rm {
n \geq ln(n) , for ~n \geq 2
\\ \therefore
\\ \frac 1n \leq \frac{1}{ln(n)}

}$$

Answer 13

what about the exponent

Answer 14

you forgot to put it there

Answer 15

\[\frac{ 1 }{ n }\le \frac{ 1 }{ (\ln n)^2 }\]

Answer 16

so the series diverges?

Answer 17

i didnt have a square in my expression

Answer 18

you had it when you multiplied both ln n's

Answer 19

the series 1/ ln(n) diverges for that reason.
but we dont know about
1 / (ln(n))^2

Answer 20

so what do we do?

Answer 21

$$

\Large \rm {
n \geq ln(n) , for ~n \geq 2
\\ \therefore
\\ \frac 1n \leq \frac{1}{ln(n)} \leq \frac{1}{(\ln n )^2}

}

$$

Answer 22

okay

Answer 23

actually this is only true for n > 2 , for n >= 3

Answer 24

$$
\Large \rm
{
n \geq \ln (n) \geq (\ln n) ^2 , ~for~ n ~\geq 3

}
$$

Answer 25

yeah that statement is true

Answer 26

now when we invert (reciprocal), the signs of the inequality is reversed.
example :

2 < 4
1/2 > 1/4

Answer 27

as long as you are positive (thats the only condition)

Answer 28

if x > y
then 1/ x < 1/y for x , y positive

Answer 29

true

Answer 30

$$
\Large \rm
{
n \geq \ln (n) \geq (\ln n) ^2 , ~for~ n ~\geq 3
\\~\\ \therefore \\ ~\\
\\\frac{1} {n} \leq \frac{1}{\ln(n)} \leq \frac{1} {(\ln n) ^2}
}
$$

Answer 31

you can show that it is true
(ln n ) ^2 > Ln (n) , for n > = 3
look at the graph
f(x) = ln(x)^2 - ln(x)
note how it dips below the x axis at first
https://www.desmos.com/calculator/pir1s5tzyn

Answer 32

thats not a proof, but it suggests it is true

Answer 33

you can take the derivative of that and show it is positive for x greater than some number between 2 and 3

Answer 34

$$
\Large \rm
{
n \geq \ln (n) \geq (\ln n) ^2 , ~for~ n ~\geq 3
\\~\\ \therefore \\ ~\\
\\\frac{1} {n} \leq \frac{1}{\ln(n)} \leq \frac{1} {(\ln n) ^2}

\\~\\ \therefore \\ ~\\
\\ \sum_{n=3}^{\infty } \frac{1} {n} \leq \sum_{n=3}^{\infty } \frac{1}{\ln(n)} \leq \sum_{n=3}^{\infty } \frac{1} {(\ln n) ^2}


}

$$

Answer 35

and you are off by one term, you can just add them through

Answer 36

but the series is n = 2 not 3 why you change it?

Answer 37

the inequality is not true for n = 2

Answer 38

its not true for n = 2