Question

Test teh following sequences for convergence or divergence. I will put teh series with teh equation box

Answer 1

Give me a miin to put up teh equation

Answer 2

what does teh mean?

Answer 3

*the

Answer 4

{1/ n ln (n)} from n=2 to infinity

Answer 5

is the ln(n) in the numerator or denominator?

Answer 6

the numerator is 1 and the denominator is n*ln(n)

Answer 7

You have to apply convergence tests.

Answer 8

we just started doing series so im not too sure how to approach the problem nor what a convergence test is

Answer 9

can u explain

Answer 10

which ones are you learning?

Answer 11

we only touched upon the nth term divergence test

Answer 12

which is?

Answer 13

if the limit n approaches infinity of a sub n is nto equal to 0, then
\[\sum_{n=1}^{\infty}\] diverges

Answer 14

Also we learned if teh limit of the nth term of {a sub n} exist, then (a sub n} converges to the limit. if it doesnt exist, then it is divergent

Answer 15

Thats basically what we learned so far

Answer 16

This series needs something more powerful.
\[\sum_{1}^{\infty}a_n \text{ converges if and only if } \sum_{1}^{\infty}2^ka_{2k}\text{ converges}\]

Answer 17

\[\sum_{k=1}^{\infty}\frac{2^k}{2^k \ln(2^k) }=\sum_{k=1}^{\infty}\frac{1}{ \ln(2^k) }=\sum_{k=1}^{\infty}\frac{1}{ k \ln(2) }=\frac{1}{\ln2}\sum_{k=1}^{\infty}\frac{1}{ k}\]

Answer 18

OMG is this the snwer

Answer 19

that last series is the harmonic series. It is also nasty to analyze (caused a lot of headaches back in the day). I think you use the same 2^k idea. But it does not converge.

Answer 20

It does diverge

Answer 21

i mean why does it diverge

Answer 22

Do you know the integral test?

Answer 23

i dont rember actually

Answer 24

\[\sum_{k=1}^{\infty}\frac{1}{k}-->\sum_{k=1}^{\infty}\frac{2^k}{2^k}=\sum_{k=1}^{\infty}1\]

Answer 25

That's the Cauchy test. The harmonic series diverges, so your original series diverges

Answer 26

um that doenst look familiar

Answer 27

First question, a very important one. You have the words "sequence" and "series" in your original question. Are these sequences or series? Their convergences are completely different.

Answer 28

I don't know all the ways to prove this stuff. But I know the ratio test, and the root test don't help.

Answer 29

Oh its a sequence

Answer 30

You mean we do not add them up!!

Answer 31

Did u add them

Answer 32

that is what the summation sign means, add up all the terms up to infinity

Answer 33

well the original question is : test the following sequences for convergence or divergence And the expresion is in brackets.

Answer 34

In that case, the sequence converges, because when n --> infinity the nth term approaches zero

Answer 35

Sorry to have dragged you through all the nasty math.

Answer 36

OH NO ....im soo sorry taht i made u do more work than necessary. its my fault

Answer 37

im soo glad u are helpingme

Answer 38

So is there any work that can be shown other than that explanation that nth term approaches 0

Answer 39

That's all you need to say

Answer 40

so does teh n=2 really matter

Answer 41

\[ \frac{1}{big number} -->0 \]

Answer 42

that makes sense

Answer 43

the ln(1)=0 , and it is not good to divide by 0, so they start with 2

Answer 44

wat does the N and Ns mean

Answer 45

or wat does that expression mean in words

Answer 46

can u help me determine the convergence/divergence of 2 more sequences after this

Answer 47

Ignore my math jargon.
A sequence converges to a number L if
\[\lim_{n \rightarrow \infty} a_n \rightarrow L\]
For example, given the sequence
\[a_n=\frac{1}{n}\]
\[\lim_{n \rightarrow \infty} \frac{1}{n} \rightarrow 0; (a_n) \rightarrow 0\]

Answer 48

oh so since it goes 2 0, it converges right?

Answer 49

Yup. It only doesn't converge if the limit doesn't exist/diverges.

Answer 50

ok cool i see. that is teh proof then. can u help me w/ 2 more???

Answer 51

did you post it

Answer 52

no i was just waitng to see if u guys dont mind helping

Answer 53

post it,