Question

check the convergence of the following:
\[\sum_{n=1}^{\infty}\frac{1}{n^{\frac{n+1}{n}}}\]
I used log to show it is divergent
\[\ln(\frac{1}{L})=\sum \ln n + \sum \frac{\ln n}{n}\]

Answer 1

how can i do it with comparison test

Answer 2

however, I'm not sure if my line of reasoning was correct considering the use of logs

Answer 3

@amistre64

Answer 4

@Kainui

Answer 5

not really a strong point of mine, ganesha seems to work these like they were 3rd grade arithmetic

Answer 6

some i struggle with comparison test
some cases it is obvious what series you need to compare, but some cases it is not so clear

Answer 7

somehow*

Answer 8

@rational you got any insights?

Answer 9

yeah ganesh is good with these stuff lol
how about my approach any missteps?

Answer 10

can we compare that with n!?

Answer 11

the question asks for comparison test!
tried other tests, they fail inclusive!

Answer 12

tried root test yet ?

Answer 13

yes failed! I my attempt was correct

Answer 14

if*

Answer 15

stumped
i bet $50 it diverges since the exponent goes to one, but i'll be damned if i see what to compare it to

Answer 16

try comparing with 1/(n*logn)

Answer 17

it is divergent 100%, problem is how to prove it is

Answer 18

so 1/nlogn is less than that expression?

Answer 19

comparison test give me hard time lol
how did you figure it out that would be good for to compare it with

Answer 20

because SUM 1/(n*ln(n)) is special and very famous
it is the middle ground between convergence and divergence of p-series

since we had a feeling that the given series diverges, and it is kinda similar to harmonic series...

Answer 21

but need to show somehow that it is less than the given series to use comparison test

Answer 22

i didn't know there was a middle ground, but i knew it diverged
but so does
\[\sum \frac{1}{n\log(n)\log(\log(n))}\] if i remember correctly

Answer 23

yeah that diverges by integral test

Answer 24

1/n diverges

`1/(n*ln(n)) diverges`

1/n^2 converges

Answer 25

hmm i see what you mean saw something like this on mit open course

Answer 26

any order greater than O(n*ln(n)) in the denominator makes the p-series converge

Answer 27

yeah i remember it from there too

Answer 28

oh what a good idea !

Answer 29

show
\[n^{\frac{n+1}{n}}<n\log(n)\] cool

Answer 30

hmm good question
induction

Answer 31

take the log lol

Answer 32

or to make ur life more simple, try limit comparison test

Answer 33

oh yeah that's a good idea

Answer 34

limit will be ugly hehe

Answer 35

\[\lim \frac{1}{n\log( n) n^{\frac{n+1}{n}}}\]

Answer 36

what would that limit go to?

Answer 37

let me try wolfram no will to do it hehe

Answer 38

goes to zero no good lol

Answer 39

yeah try comparison oly as satallite said

Answer 40

need to show that inequality
i will try to see if that holds for some n or any n

Answer 41

not good for any n

Answer 42

n>=5 seems where it is true

Answer 43

try this

For all positive \(n\), we have \(\large \sqrt[n]{n} \lt 2\)

Answer 44

what is the goal?

Answer 45

\[\large {\sqrt[n]{n} \lt 2 \implies \frac{1}{\sqrt[n]{n}}\gt \frac{1}{2} \implies \frac{1}{n*\sqrt[n]{n}}\gt \frac{1}{2n} }\]

Answer 46

i see
1/2n diverges

Answer 47

that will do yeah

Answer 48

that's pretty good actually :)

Answer 49

great job man :)

Answer 50

np :)
is it easy to show that \(\sqrt[n]{n}\) is less than 2

Answer 51

hmm let me do that lol
direct proof

Answer 52

i took it to \[\ln \frac{n}{2^n}<0\]
am i deviating

Answer 53

since n/2^n <1 for any n
then ln(n/2^n)<0
have to start from here and go backwards

Answer 54

sounds good?

Answer 55

thats pretty clever!

Answer 56

:)

Answer 57

then we are done :)
thanks a lot feel good this is solved lol

Answer 58

:D

Answer 59

that ln(1/L) thingy really worked ?

Answer 60

im referring to the main q
http://gyazo.com/a9621914c1299a48d614cbed5c2888e8

Answer 61

will the left hand side is divergent
but the reasoning lacks something hehe

Answer 62

Okay that looked bit unorthodox haha! but its a good idea

Answer 63

just tried a bunch of things lol
no harm in exploring:)