Question

Test the series for convergence or divergence.

Answer 1

\[\sum_{k=1}^{\infty}\frac{ k*\ln(k) }{ (k+1)^3 }\]

Answer 2

Looks like the comparison test is needed, but I'm pretty weak at using comparison test. Help is appreciated.

Answer 3

lnk<k for k>1

Answer 4

k>=1

Answer 5

therefore
k^2
---
k^3
is right on the border, our is less thn this so it converges

Answer 6

What you are saying that it less than the Harmonic Series 1/k which is divergent, it does not mean that the original series is convergent

Answer 7

yes i know 1/k is divergent

Answer 8

but we know ourseries is less than 1/k

Answer 9

hmm is that not good enough

Answer 10

the top was k^2, but we know the power is slightly less than 2

Answer 11

well i shouldnt say slightly less than 2, it is strictly less than 2

Answer 12

so this power p is more than 1
1/k^p,

Answer 13

How do you show that is less than 1/k^p where p>1

Answer 14

The best way is to use the integral test and notice that
\[
\int_1^{\infty } \frac{x \log (x)}{(x+1)^3} \, dx=\frac{1}{4} (1+\log (4))
\]

Answer 15

oh another solution ln(x) < sqrt(x)

Answer 16

The series is convergent by the integral test

Answer 17

Yes, your last argument will do

Answer 18

Your last argument is easier to use than the integral test

Answer 19

yeah, but what will we do about the k+1
i just realized i was claiming
1/k < 1/(k+1), which is wrong

Answer 20

unless we do

Answer 21

k^(1.5) (k+1)^1.5
------ < ---------
(k+1)^3 (k+1)^3

and we know that
1/x^1.5
is convergent for x<inf

Answer 22

Yes

Answer 23

\[
\frac{k \ln(k)}{(k+1)^3}<\frac{k^{3/2}} {k^3}\approx \frac 1 {k^{3/2}}

\]
for k greater or equal to one

Answer 24

x^0.5,lnx
raise both by e
e^(x/2) , x
if u look at x = 1
e^0.5 > 1
and we know thatthe slope of e^(x/2), will be greater than 1 after that too, so this exponential is more than x
which means x^0.5 > lnx for all x=1 and more
(note its actually more right from 0)

Answer 25

thats to show sqrt x > ln x