Question

Doubt in convergence of sum of different series

Answer 1

The question is :

What is the range of values of x for which :

\[\large{\sum{(a + b + c)}}\] is convergent.

Answer 2

a, b and c are some series...

Now, I solved this question in following way.

1. I proved that series b is convergent for all values of x.

2. Then, I said that for the sum to be convergent, series (a+c) must also be convergent

Answer 3

so the sum an + bn + cn

Answer 4

3. Then using power series formula, I evaluated the set of those values of x.

Now, my answer matches with the correct answer. But, my teacher says that I am wrong.

Answer 5

the radius of convergence will be the intersection of the radii of convergence of an, bn, and cn.

Answer 6

@perl no no... a, b and c are not constants. They are some functions in terms of x. For saving time, I have not mentioned the exact functions.

Answer 7

@perl exactly!!! That is what I did, but my teacher says I am wrong. I would like to know your views on this

Answer 8

are the series, an, bn, and cn given?

Answer 9

yes

Answer 10

i would need to see them

Answer 11

Does your teacher give a reason why you're wrong, maybe you should talk with your teacher and it's just a simple mistake.

Answer 12

Perhaps some step in your proof is incorrect, even though you ended up at the correct answer your argument was flawed.

Answer 13

I know if I was an algebra teacher and someone said \[\LARGE 3^2*3^2=3^{2*2}=81\] I would say they were wrong lol.

Answer 14

Hi @Kainui , I asked my instructor and he said that you can't do this.... when I asked what is wrong with this, he just repeats himself and says, it is not the correct way of solving questions...

That is why, I would like to know what is wrong with my method...

In fact, in thomas calculus, a question was solved using this approach... :D

This question has 18 credit points, those who only wrote correct answer (i.e. the correct range got 2) but after complete explanation and correct answer, I got zero ;)

I gave it for rechecking and still the same marks... It is not that I want marks. Its just that I want an explanation for why this approach is wrong

Answer 15

@perl, sure... Give me a few minutes please

Answer 16

The only explanation that seems to make sense to me is that you are smarter than your teacher.

Answer 17

:D If I say this to him... he is going to kill me ;)

Answer 18

Hahaha I am only telling you so that you don't kill yourself trying to reason out why you are wrong when you are perfectly right.

Answer 19

Okay here is the exact question:

Find the positive values of x for which the following series converges:

\[\large{\sum_{n=1}^{\infty}[\cfrac{x^{3n-2}}{(3n-2)!} + \cfrac{1}{(3n-1)^{x+1}} + x^{3n}}]\]

Answer 20

First, I proved that \(\large{\cfrac{1}{(3n-1)^{x+1}}}\) converges for all positive values of x.

Then, I mentioned that sum of rest two terms must also converge for the whole sum to be convergent

Answer 21

I think the third term there only converges for x^3<1 since it's a geometric series.

Answer 22

The first term converges for all x since it's only every third term of the power series for e^x which already converges, so taking away some of the terms of that will only make it converge to smaller numbers.

Just throwing out facts and ideas to help, keep explaining what you did though and I'll see what I can do.

Answer 23

For convergence of first and third terms,

\[\large \lim_{n \rightarrow \infty}|{\cfrac{{u_{n+1}}}{u_{n}}| < 1}\]

Answer 24

Even though this isn't the limit as you get infinitely larger, if you did pick x=2 and looked at the first two terms of the summation, n=1 and n=2 \[\LARGE \left| \frac{2^6}{2^3} \right| \ge1\] So you know that is false.

Answer 25

\[\large{\implies \lim_{n \rightarrow \infty} |\cfrac{x^{3n+3} + \cfrac{x^{3n+1}}{(3n+1)!}}{x^{3n} + \cfrac{x^{3n-2}}{(3n-2)!}}| < 1}\]

\[\large{\implies\lim_{n \rightarrow \infty} |x^3\cfrac{x^2 + \cfrac{1}{(3n+1)!}}{x^2 + \cfrac{1}{(3n-2)!}}| < 1}\]

\[\large{\implies\lim_{n \rightarrow \infty} |x^3| < 1}\]

\[\large{\implies \lim_{n \rightarrow \infty} |x| < 1}\]

\[\large{\implies -1 < x < 1}\]

Since, x is positive, thus:

\[\large{0<x<1}\]

Answer 26

@Kainui , we are given that series is convergent, so we will have assume that it is less than 1 and using which we can find out the correct set of values of x.

And also x = 2 > 1, which will not satisfy the convergence condition (as you mentioned)

Answer 27

Sorry I thought earlier you said the radius of convergence was 2 that's why I wrote that.

Answer 28

Sorry about that :)

Answer 29

What do you think @Kainui and @perl ?

Answer 30

but x is not zero (it is given that x is positive)

Answer 31

oh nevermind, you wrote x > )

Answer 32

x > 0

Answer 33

:)

Answer 34

ok so we can concentrate on the first and third series

Answer 35

yeah

Answer 36

i would prove the first series converges separately, and the radius

Answer 37

and your teacher said the answer is wrong, that you gave? what was your answer

Answer 38

My final answer was the required values of x is (0,1) and this is the correct answer according to him also, but he says that your method is wrong. He doesn't explain why ...

Answer 39

well it seems because you need x > 0 in order that the middle series does not diverge.
and you have |x | <1 because of the geometric series (for the third series).

Answer 40

so putting these two facts together, you get it converges on (0,1)

Answer 41

yes

Answer 42

he said that's wrong?

Answer 43

no he says that the answer is correct but the approach is wrong...

Answer 44

What do you think ?

Answer 45

well the sum of two converging series is converging

Answer 46

yep

Answer 47

I think you can find the interval of convergence for each series separately. then find the 'intersection'
so the first series converges on (-oo,oo).
second series converges on (0,oo)
third series converges on (-1,1)

so the intersection is (0,1)

sorry, i dont know what else you could do .

Answer 48

and since each other series converges separately, the sum of the series converges

Answer 49

I agree that there are different methods of solving this question, but I just don't get what is wrong with my method ;)

But still, thanks a lot for your response @perl and @Kainui

Answer 50

i would need more information from your teacher. maybe he didnt like how you tried to convergence of two series in one ?

Answer 51

it looks like you combined two series in one

Answer 52

you found the convergence of series 2 and series 3 , you did it one shot. i think you should do them separately

Answer 53

that could be a problem, if you have series that somehow cancel each other out

Answer 54

That may be the case... even then I will ask him again

Answer 55

What I am thinking
if you can show that series an converges and series bn converges separately
THEN
series (an +bn) converges.
This is a theorem (I checked).

The converse may not be true

Answer 56

the converse may not be true generally

Answer 57

Yes you are correct on that.

There are only 2 possibilities for the sum of 2 series to be convergent:

1. Both convergent
2. Both divergent

Answer 58

this is a sum of three series, not two series

Answer 59

Yes but I have transformed into sum of 2 series - 1. a+c 2. b

Answer 60

but you have to follow the theorem of a sum of series! the hypothesis says

Answer 61

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&sqi=2&ved=0CDIQFjAC&url=https%3A%2F%2Fmath.la.asu.edu%2F~kawski%2Fclasses%2Fmat300%2Fhandouts%2Fconvsumprod.pdf&ei=70pPVMilF8b9yQSCv4GoDQ&usg=AFQjCNFYE7RGlHTIvEmQrdduyySquqN-4Q&sig2=-lOShbGzUSwJgh_VTFA_kg&bvm=bv.77880786,d.aWw

Answer 62

I did not say that a and c must be convergent, I said that (a+c) must be convergent

Answer 63

im not sure if thats true

Answer 64

theorem:
The sum of two converging sequences converges.
(so you have to know the series are convergent seperately before you take the sum)

Answer 65

Correct.

But, if you are given that sum of 2 series is convergent and one of the series is convergent, then other must also be convergent

Answer 66

Because if it is divergent, then, the sum will become divergent

Answer 67

yes, but are you given that in this problem?

Answer 68

yes

Answer 69

i didnt see that given

Answer 70

1. Find the positive values of x for which the following series converges:

So, I have to assume that the whole sum must be convergent.

2. I proved that b is convergent.

So, sum of 2 series (a+c) and b is convergent, b is convergent so (a+c) should be convergent

Answer 71

ok, checking

Answer 72

Though, it does not mean that a and c will be convergent but that is not necessary for the problem

Answer 73

sure :)

Answer 74

i wouldnt go about assuming the whole series converges. you can find that out as you do the problem. there may be no solution

Answer 75

But we have to find out the values of x for which the whole series converge.

Answer 76

right, and you can do that by finding the interval of convergence seperately for the three series.

Answer 77

Yes, that is also a possible approach

Answer 78

its possible there are no x values for which there is a solution. so thats kind of a big assumption.

Answer 79

My instructor agrees on this that we can assume that whole series is convergent because many students used this fact and then used ratio test to find out the values of x and he gave them full marks.

Answer 80

sorry, its possible that there is no x values for which the whole series is convergent

Answer 81

hmmm, then there could be another problem. i dont see it :)

Answer 82

I will ask my instructor again... :)

Thanks again

Answer 83

let me look over your arithmetic, one sec

Answer 84

sure

Answer 85

i mean your algebra

Answer 86

bear with me

Answer 87

Sure :)

Answer 88

i have some issue with your algebra

Answer 89

in your limit

Answer 90

did you factor out x^(3n) ?