Question

How to prove that a series is convergent by using Cauchy Convergence Criterion for Series?

Answer 1

define the criterion

Answer 2

im thining its: for e > 0 and m,n > N |xm - xn| < e

Answer 3

The series \(\sum_{n=1}^\infty a_n\) converges if and only if for every \(\epsilon > 0\) there exists \(N_\epsilon> N\) such that for every \(n > m > N_\epsilon\) we have \(|a_{m+1} + a_{m+2} + + a_n| < \epsilon\)

Answer 4

Cauchy’s criterion.

The sequence xn converges if and only if for every e > 0 there exists a natural number N

such that |xn − xm| < e whenever n, m > N.

Answer 5

im not sure im familiar with your definition of it

Answer 6

For Cauchy Convergence Criterion, there are two versions: one is for sequence, and the other one is for series. Yours is for sequence, mine is for series.

Answer 7

hmm, then that would explain the differences :)

Answer 8

so, what is your series that you are trying to cauchy?

Answer 9

Well, I am not sure if I can post my homework problem. Let me just make up a similar one... \(\sum_{n=1}^\infty\frac{1}{n^2-5}\)

Answer 10

so now we want to make a partial sum, it looks like

Answer 11

Yes.

Answer 12

will show u ltr (after 3,4 hours ), if u still need it hehehe
its much easy than u think :O

Answer 13

if you have some guidance, feel free to participate :)

Answer 14

yeah basically you're just proving the sequence of partial sums converges

Answer 15

@ikram002p
What is the main idea of your approach?

Answer 16

\[\text{sequence of partial sums converges} \iff \text{series converges}\]

Answer 17

comparison using p-series is more natural for proving 1/(n^2-5) series though

Answer 18

Here's what I've got so far:
\(|S_n-S_{n-1}| = a_n\)
For n>m, we can use triangle inequality to obtain
\(|S_n-S_m| = |S_n - S_{n-1} + S_{n-1} + S_{n-2}+ ... +S_{m-2}-S_{m-1}+S_{m-1} -S_m|\)
\(\le |S_n - S_{n-1}| + |S_{n-1} + S_{n-2}|+... +|S_{m-2}-S_{m-1}|+|S_{m-1} -S_m|\)
\(=|a_n|+|a_{n-1} + ... + |a_{m-2}|+|a_{m-1}|\)
Then... lost...

Answer 19

Apart from proving it like that, I am trying to show that \(a_n\) is a montone sequence, so I would have \(a_n\) = convergent => \(a_n\) = cauchy, then done.
But it doesn't seem to be a monotone sequence due to the first term...

Answer 20

u chosed n as n-1 and n as n , lets if its work , u need to expand a bit to make useful of scoping ,
it is monotone but u need to follow steps to prove using cauchy , hmm saying monotone is not enough , lets walk through the Def , again ok ?

Answer 21

well you can always get rid of first few terms with their partial sum :
\[\large \sum\limits_{n=1}^{\infty}\frac{1}{n^2-5} = S_m+ \sum\limits_{n=m+1}^{\infty}\frac{1}{n^2-5} \]

Answer 22

but we should note its monotone after n=2 hmm buts thats also work

Answer 23

If it is monotone and bounded, then it is convergent. A convergent sequence is a Cauchy sequence. Then we can use it to show that its series is also convergent by the Cauchy Convergence Criterion for series.
Is it right?

Answer 24

The trickiest part is... in my lecture notes, we have to include the case when n=1 for a sequence to be a monotone sequence.

Answer 25

ok go like this ,
chose as u said m=n+1
then m>n
|s_(n+1) - s_(n) |= |x_1+x_2 +( the rest go with scoping ) |

Answer 26

telescoping ?

Answer 27

@ganeshie8 How and what to telescope?

Answer 28

no, im asking ikram if she has meant to use telescoping for the sum inside absolute bars

Answer 29

yeah hehe but scope means chicking , for our luck its telescope here xD

Answer 30

checking*

Answer 31

What is x_1?

Answer 32

very interesting

Answer 33

its the value of x_n when u apply 1
like
x_1= 1/(1)^2-5=1/-4
x_2=1/(2)^2-5 + 1/-4
x_n=.....

Answer 34

Then no, |s_(n+1) - s_(n) | is not equal to |x_1+x_2 +...|, but ||s_(n+1) - s_(n) | = x_(n+1)

Answer 35

hehe only in this case :3

Answer 36

u've already chosed m=n+1

Answer 37

If m = n+1, then it violates my assumption that n>m

Answer 38

i said m>n :O

Answer 39

its ok its just a matter of ordering , i worked on
m>n>N(epsilon )
u worked on
n>m>N(epsilon )
i dint note that sry :3

Answer 40

Can you please explain this part then?
|s_(n+1) - s_(n) |= |x_1+x_2 +( the rest go with scoping ) |

Answer 41

\(S_{n+1} = \sum_{i=1}^{n+1}x_i = x_1+x_2+...+x_n+x_{n+1}\)
\(S_{n} = \sum_{i=1}^{n}x_i = x_1+x_2+...+x_n\)
\(S_{n+1}-S_{n} = x_{n+1}\)
This is what I get

Answer 42

oh wait , thats right xD

Answer 43

And I don't see how much it differs from my working, despite that I have done more than that.

Answer 44

ugh i told u u worked right from the start just expand xD
then
1/(n+1)^2-5 = 1/(n^2+2n-4) <2

Answer 45

How do you get this: 1/(n^2+2n-4) <2?

Answer 46

its monotone right ?
sup of -1 |-1|=1
take any sup value >1 and make it epsilon
u could chose \(\epsilon_0 + 1\)

Answer 47

sup= supremum

Answer 48

No, it is NOT monotone

Answer 49

For monotone, it is either increasing or decreasing for ALL \(n\in \mathbb{N}\)

Answer 50

first few terms should not stop you from applying convergence tests
its the tail that matters

Answer 51

not series only x_ n+1
-1, 1/4, ....
xD
-1 not in terms and also 2 > -1 xD

Answer 52

-1/4, -1, 1/4
Not increasing nor decreasing

Answer 53

I understand that the first few terms doesn't matter for the convergence of the whole sequence/series, but what bugs me is when I put it in the proof, it is not that persuasive at all, especially when you claim it is monotone while it is not.

Answer 54

i might made a typo
x_n+1=1/(n^2+2n-4) =-1,1/4,1/11., ...
how about now monotone or what O.O

Answer 55

you just x_n+1

@ganeshie8 , hehe anything better than cauchy , even thought its the fastest way to say its converge xD

Answer 56

Perhaps not. Maybe the fastest way to say it converge is by the comparison test.

Answer 57

that also my fav :O
but cachy is for prof :3

Answer 58

x_n+1=1/(n^2+2n-4) =-1,1/4,1/11., ...
^ still not monotone

Answer 59

anyay say its monotone for n>1 ,
thus take any value >1 (since first term is 1)
and make sure it goes will with other terms

the idea is not about monotone or not , its about finding epsilon that make any any value of the sequence x_n+1 < epsilon

Answer 60

do u have a problem by saying for n>1 its monotone ?

Answer 61

when u start it from 1/4,.....
(PS we are working on x_n+1 = 1/n^2+2n-5) )

Answer 62

take sup =1/4 for example
or 1
let it be 1

Answer 63

No, if I can say it is monotone for n>1 and ignore the case when n=1, I know how how to finish this problem.
Just show it is monotone + bounded => convergence => Cauchy => done

Answer 64

ugh

Answer 65

u said u wanna CC for series

Answer 66

Yes, but CC for series came from CC for sequence

Answer 67

if u wanna show its monotone + bounded its a lema
like in other question , if its been telling u show its cauchy then u can say its bounded + monotone done

Answer 68

haha i dont know if u see a difference xD
but u will in first exam^_^
be carful :P

Answer 69

but RolyPoly is particular about the first two terms

Answer 70

he/she wants to use the sequence from 1->infinity

Answer 71

so ?

Answer 72

so the given sequence as-it-is is not monotone

Answer 73

because it has few bad terms in the start

Answer 74

lol im fine skipping past first few terms if my only goal is to test convergence

Answer 75

but if you want to write a neat proof, you will need to think all the corners

Answer 76

^Exactly

Answer 77

yes

Answer 78

however its ur choice at the end :)

Answer 79

im getting late for dinner, brb

Answer 80

Are there no ways to fix this problem though?

Answer 81

leme tag @nerdguy2535 before i leave

Answer 82

my way is like this :-
1- reduce the series into limited sequences
i like m=n+1 ,m>n>=N(epsilon )
|s_m-s_n|= x_(n+1) ( all ur work )

2- according to CC for series
for all epsilon >0 , there exist N(epsilon) in n
such that if m>n>=N(epsilon), then
|s_m-s_n| =|...| <epsilon

so ,
|x_(n+1)|= |1/n^2+2n-4| =|1,1/4,1/11,.... 1/n^2+2n-4| < epsilon
take \( \epsilon = \epsilon_0 + 1 , \epsilon_0>1 \)

hehehe xD
@nerdguy2535 help lol

Answer 83

Whats the exact question? Trying to show that the series: $$\sum_{n=1}^\infty\frac{1}{n^2-5}$$converges using the Cauchy Criterion?

Answer 84

yeahh

Answer 85

I think I saw this somewhere earlier in the discussion. If \(n>m\), and \(S_n=\sum_{k=1}^n \frac{1}{k^2-5}\)then: $$\left|S_n-S_m\right|=\left|a_{m+1}+a_{m+2}+\cdots +a_n\right|$$$$\le |a_{m+1}|+|a_{m+2}|+\cdots +|a_n|$$using the triangle inequality. Is everyone up to here?

Answer 86

waiting u hehe

Answer 87

yes

Answer 88

and Joseph went off xD

Answer 89

(i'm kinda in a hurry so I dont have too much time unfortunately).

The idea is that since \(k^2-5<(k+1)^2-5\) for all k, we have: $$\frac{1}{(k+1)^2-5}<\frac{1}{k^2-5}.$$ So: $$|a_{m+1}|+\cdots +|a_n|\le (n-m)\cdot |a_{m+1}|.$$

Answer 90

Try and show that as m goes to infinity, that $$(n-m)\cdot |a_{m+1}|\rightarrow 0$$

Answer 91

Call n-m=k for some fixed k (since n>m).

Answer 92

So you have: $$\left|\frac{k}{(m+1)^2-5}\right|,$$which you want to show goes to zero as m goes to infinity.

Answer 93

Does that make sense?

Answer 94

cool :O

Answer 95

By forcing n>m, it suffices to just take m to go to infinite, because it will also take n to infinity. n=m+k after all.

Answer 96

\[ \left| \frac{1}{n^2-5} + \dots + \frac{1}{m^2-5} \right | < \left| \frac{1}{n^2} + \dots + \frac{1}{m^2} \right | \le \left| \frac 1 {n^2} + \frac{1}{n(n+1)} + \dots +\frac {1}{m(m-1)}\right| \\ = \left| \frac 1 {n^2} + \frac 1 n - \frac {1}{m}\right| < \frac{1}{n^2} + \frac 1 n = \epsilon \]

Answer 97

Ah! I'll try to work on it later. It's too late in my place and I have to go. Sorry!