Question

Converge or diverge?
\[\sum_{n=0}^\infty \dfrac{\sqrt{n+1}}{n^2-3n+1}\]
Please, help

Answer 1

@rational

Answer 2

are you allowed to use comparison test ?

Answer 3

Yes

Answer 4

.

Answer 5

For \(n\gt 4\) we have \[\dfrac{\sqrt{n+1}}{n^2-3n+1}\lt \dfrac{\sqrt{n+1}}{n^2-3n-4}=\dfrac{\sqrt{n+1}}{(n+1)(n-4)}\\=\dfrac{1}{\sqrt{n+1}(n-4)}\lt \dfrac{1}{\sqrt{n-4}(n-4)}=\dfrac{1}{(n-4)^{3/2}}\]

Since \(\frac{3}{2}\gt 1\), the series converges by p-series test.

Answer 6

If it is so, we depend on n is not 4, right? while the original one is not. How to argue? because original one gave us the rank from 0 to infinitive

Answer 7

Easy, split the series into two :
\[\sum_{n=0}^\infty \dfrac{\sqrt{n+1}}{n^2-3n+1}=\sum_{n=0}^4 \dfrac{\sqrt{n+1}}{n^2-3n+1}+\sum_{n=5}^\infty \dfrac{\sqrt{n+1}}{n^2-3n+1}\]

the first series on right hand side evaluates to some number, and the second series converges by comparison with p-series. so the overall series converges.

Answer 8

I got it. Thank you so much. Is there any way for my second one?

Answer 9

is there a typo in second series
the index variable "n" doesn't agree with the varibale in expression "k"

Answer 10

I am sorry, it is k =1 to infinitive

Answer 11

\[\sum_{k=1}^\infty \dfrac{log(k+1)-log k}{tan^{-1}(2/k)}\]

Answer 12

try limit test

Answer 13

\[\lim\limits_{k\to\infty} \dfrac{log(k+1)-log k}{tan^{-1}(2/k)}= ?\]

Answer 14

I know ln (1+1/k) goes to 0 if k goes to infinitive. How to argue with denominator?

Answer 15

lhopital allowed ?

Answer 16

Ok with first look,convergence

Answer 17

May be!! I am allowed to use whatever.

Answer 18

use lhopital rule then

Answer 19

But it turns out to convergence when the answer from the book is divergence

Answer 20

how do you say it converges ?

Answer 21

Basically cuz im talking about the one on the question xD

Answer 22

yes the first series converges, im asking loser how he is saying the second series with logs converges

Answer 23

numerator: (log (1+1/k))'= 0
denominator (tan^-1 (2/k)' = -2/(k^2+4)

Answer 24

ok, it is divergent. Thank you

Answer 25

Give me the trick, please. How can you quickly figure out which method we should apply?

Answer 26

why is it divergent ?

Answer 27

as k --> infinitive, lim is undefined

Answer 28

wolfram says the limit is 1/2

http://www.wolframalpha.com/input/?i=lim+%5Cdfrac%7Blog%28k%2B1%29-log+k%7D%7Btan%5E%7B-1%7D%282%2Fk%29%7D+as+k%5Cto%5Cinfty

Answer 29

the series diverges by limit test because the limit is NOT 0.

Answer 30

Limit test :

\(\lim\limits_{n\to\infty}a_n\ne0 ~~\implies ~~~\sum\limits_{n=1}^{\infty} a_n\) diverges

Answer 31

MY God!! I confused everything.

Answer 32

thats okay, so basically one always tries "limit test" first,
if the limit is not 0, then the series diverges.

Answer 33

here is my preferred order of tests for series with positive terms :

1) limit test
2) ratio test
3) comparison test
4) limit comparison test
5) root test
6) others

Answer 34

You mean I have to take one by one, if limit test fail, then try ratio, then....., right?

Answer 35

upto you, if limit test is inconclusive, then i generally think of ratio test next

but thats not a strict rule, i sometimes try comparison test directly for series like : \(\sum\limits_{n=0}^{\infty}\dfrac{1}{n^2+1}\)

Answer 36

i like comparison test more but ratio test gets the work done quickly for most series in textbooks

Answer 37

For comparison test, we need manipulate the original series to get the familiar form, how to quickly figure it out? Like the first one of my problem.

Answer 38

If I use comparison instead of p-series, how can I find \(b_n\)? the series I want to compare to \(a_n\)

Answer 39

practice i guess, i don't have any good tips on recognizing when to use which test...

Answer 40

Ok, Thanks a lot. If it is "practice", I have to. No choice, right?

Answer 41

the first series is worked using "comparison test", we compared it with the familiar p-series :

Since \(\dfrac{\sqrt{n+1}}{n^2-3n+1} \lt\dfrac{1}{(n-4)^{3/2}}\) for \(n\gt 4\), and the series \(\sum\limits_{n=5}^{\infty}\dfrac{1}{(n-4)^{3/2}}\) converges by p-series test,

the actual series \(\sum\limits_{n=5}^{\infty}\dfrac{\sqrt{n+1}}{n^2-3n+1} \) converges by "comparison test".

Answer 42

keep in mind it is "comparison test", not p-series test.

p-series is the series we compared against

Answer 43

What if I use comparison test with \(b_n= \dfrac{1}{(n+1)^{3/2}}\)

Answer 44

do you mean "limit comparison test" ?

Answer 45

that works equally well!

Answer 46

yes

Answer 47

Thank you so much.

Answer 48

np:)