Please help me with yet ANOTHER infinite series question


by skyturnred
Tags: infinite, series
skyturnred
skyturnred is offline
#1
Feb15-12, 01:54 PM
P: 116
1. The problem statement, all variables and given/known data

Ugh really getting frustrated at not being able to keep up with math..

Find the set of x for which the series converges AND find the sum of the series at these values.

s=[itex]\sum^{k=infinity}_{k=0}[/itex] [itex]\frac{(x+7)^{k}}{3^{k}}[/itex]

2. Relevant equations



3. The attempt at a solution

OK, by using the ratio test I find the set to be x=(-10,-4). I entered this in, and this is correct. But I am very confused about the second part of the question.. starting with the wording.

Is it asking me to find the sum at x=-10, THEN the sum at x=-4, and then add them? I ask this because there is only ONE field in which to input the answer.

But regardless of whether it is asking that, I am finding it difficult to find the sum when x=-10.

I get the sum of (-1)[itex]^{k}[/itex] from k=0 to k=n as n approaches infinity. So I get the sum of

s=1-1+1-1+1+...(-1)[itex]^{n-1}[/itex]+(-1)[itex]^{n}[/itex]

My initial thought was that s=0, because all the terms would cancel out. Then I realized, it could also equal 1 because as n approaches infinity, the terms would always change between 1 and -1. So I couldn't decide between the two.

But THEN I entered it into wolfram alpha, and it says that the series is equal to 1/2! How does this even make sense? Can someone explain please?

But I am very confused AGAIN for the following: When I try x=-4, I get the sum of 1+1+1+1+1+... n times (so it should be infinity). And if the original question is asking to add these series together, then it shouldn't matter whether the sum when x=-10 is 0, 1, or even 1/2, because any of those values plus infinity is equal to infinity. But when I input this into the answer area, I get it wrong. What am I doing wrong?

Thanks.
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Ray Vickson
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#2
Feb15-12, 02:16 PM
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Quote Quote by skyturnred View Post
1. The problem statement, all variables and given/known data

Ugh really getting frustrated at not being able to keep up with math..

Find the set of x for which the series converges AND find the sum of the series at these values.

s=[itex]\sum^{k=infinity}_{k=0}[/itex] [itex]\frac{(x+7)^{k}}{3^{k}}[/itex]

2. Relevant equations



3. The attempt at a solution

OK, by using the ratio test I find the set to be x=(-10,-4). I entered this in, and this is correct. But I am very confused about the second part of the question.. starting with the wording.

Is it asking me to find the sum at x=-10, THEN the sum at x=-4, and then add them? I ask this because there is only ONE field in which to input the answer.

But regardless of whether it is asking that, I am finding it difficult to find the sum when x=-10.

I get the sum of (-1)[itex]^{k}[/itex] from k=0 to k=n as n approaches infinity. So I get the sum of

s=1-1+1-1+1+...(-1)[itex]^{n-1}[/itex]+(-1)[itex]^{n}[/itex]

My initial thought was that s=0, because all the terms would cancel out. Then I realized, it could also equal 1 because as n approaches infinity, the terms would always change between 1 and -1. So I couldn't decide between the two.

But THEN I entered it into wolfram alpha, and it says that the series is equal to 1/2! How does this even make sense? Can someone explain please?

But I am very confused AGAIN for the following: When I try x=-4, I get the sum of 1+1+1+1+1+... n times (so it should be infinity). And if the original question is asking to add these series together, then it shouldn't matter whether the sum when x=-10 is 0, 1, or even 1/2, because any of those values plus infinity is equal to infinity. But when I input this into the answer area, I get it wrong. What am I doing wrong?

Thanks.
You have a sum of the form [tex]S(x) = \sum_{k=1}^\infty y^k, \mbox{ where } y = \frac{x+7}{3}. [/tex]
Surely you have seen such sums hundreds of times already.

To me the question seems perfectly clear: evaluate S(x) for any value of x in the convergence region. It does not ask you to sum S(x) for different values of x, because that would make no sense at all: summing S(x) for a continuous x is not a defined operation (although integrating S(x) would be). Certainly you should never attempt to add the values of S at the endpoints of the convergence region because these values may not exist: convergence regions of infinite series are open intervals, and the series might (usually does) diverge at the endpoints of these regions.

PS: in tex the command for ∞ is "\infty", not "\infinity" or "infinity".

RGV
skyturnred
skyturnred is offline
#3
Feb15-12, 03:14 PM
P: 116
Quote Quote by Ray Vickson View Post
You have a sum of the form [tex]S(x) = \sum_{k=1}^\infty y^k, \mbox{ where } y = \frac{x+7}{3}. [/tex]
Surely you have seen such sums hundreds of times already.
I have not seen hundreds of such sums or hundreds of ANY infinite sums for that matter. I am attending a first year calculus course in which we only just started sums last friday. I would not post a question to physicsforums unless I cannot figure it out.

Quote Quote by Ray Vickson View Post
To me the question seems perfectly clear: evaluate S(x) for any value of x in the convergence region. It does not ask you to sum S(x) for different values of x, because that would make no sense at all: summing S(x) for a continuous x is not a defined operation (although integrating S(x) would be).
With all due respect, you must be wrong here. There is a single answer field in which to input a response. Since the answer changes as x changes from -10 and -4, this is clearly not what is expected. This is backed by a few of my fellow engineering majors who have tried what you mentioned above, and had the wrong answer.

Quote Quote by Ray Vickson View Post
PS: in tex the command for ∞ is "\infty", not "\infinity" or "infinity".
Im sorry for this simple mistake that obviously did not change the question in any way nor could possibly have annoyed anyone with normal tolerance/ patience levels.

Thanks anyways

vela
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#4
Feb15-12, 03:51 PM
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Please help me with yet ANOTHER infinite series question


Quote Quote by skyturnred View Post
I have not seen hundreds of such sums or hundreds of ANY infinite sums for that matter. I am attending a first year calculus course in which we only just started sums last friday. I would not post a question to physicsforums unless I cannot figure it out.
It's a geometric series, which, in the US at least, is covered in high school math and probably in junior high. You've likely seen it before but have since forgotten about it.

With all due respect, you must be wrong here. There is a single answer field in which to input a response. Since the answer changes as x changes from -10 and -4, this is clearly not what is expected. This is backed by a few of my fellow engineering majors who have tried what you mentioned above, and had the wrong answer.
The question is asking you for the sum of the series at values of x for which it converges. There's an infinite number of possible values, so obviously you're supposed to come up with an expression S(x) that depends on x. Your interpretation can't possibly be right because S(-10) and S(-4) don't exist.

But regardless of whether it is asking that, I am finding it difficult to find the sum when x=-10.

I get the sum of (-1)k from k=0 to k=n as n approaches infinity. So I get the sum of

s=1-1+1-1+1+...(-1)n−1+(-1)n

My initial thought was that s=0, because all the terms would cancel out. Then I realized, it could also equal 1 because as n approaches infinity, the terms would always change between 1 and -1. So I couldn't decide between the two.
When you say an infinite series converges, it means the sequence of partial sums sn converges. In this case, you find sn flips back and forth between 1 and 0 (not -1), so the series doesn't converge.
Ray Vickson
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#5
Feb15-12, 04:35 PM
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Why would you suppose I was imptatient or annoyed?YOU used Tex, but got the infinity sign wrong, so I thought you might like information on how to do it right. As for never having seen such a series before: if you haven't, I am truly surprised, especially in a "Calculus and beyond" forum.

RGV
skyturnred
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#6
Feb15-12, 04:49 PM
P: 116
Quote Quote by Ray Vickson View Post
As for never having seen such a series before: if you haven't, I am truly surprised, especially in a "Calculus and beyond" forum.

RGV
Quote Quote by vela View Post
It's a geometric series, which, in the US at least, is covered in high school math and probably in junior high. You've likely seen it before but have since forgotten about it.
I apologize for being rude. Although it's no excuse, I have just been getting increasingly frustrated at simply not understanding what many others in my faculty seem to be able to understand in their sleep.

Also, I went to a terrible high school. Almost none of my graduating class went on to post-secondary education. I'm almost positive we didn't do series. I asked my friend who also went to my high school and he says the same thing.

Quote Quote by vela View Post

When you say an infinite series converges, it means the sequence of partial sums sn converges. In this case, you find sn flips back and forth between 1 and 0 (not -1), so the series doesn't converge.
Thanks for letting me know this, I wasn't quite understanding exactly what divergence meant. I thought a sum HAD to go off to infinity for it to be "divergent," but I guess it is considered "divergent" if it doesn't converge as well.
vela
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Feb15-12, 05:10 PM
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Quote Quote by skyturnred View Post
Also, I went to a terrible high school. Almost none of my graduating class went on to post-secondary education. I'm almost positive we didn't do series. I asked my friend who also went to my high school and he says the same thing.
Unfortunately, it wouldn't be terribly surprising if that is indeed the case. Fortunately, the geometric series isn't terribly difficult to master even if you are seeing it for the first time.


Thanks for letting me know this, I wasn't quite understanding exactly what divergence meant. I thought a sum HAD to go off to infinity for it to be "divergent," but I guess it is considered "divergent" if it doesn't converge as well.
Yeah, that's a common mistake. It seems like diverge should mean it goes off to infinity, but in the context of series, it simply means doesn't converge.
Ray Vickson
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Feb15-12, 05:25 PM
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Quote Quote by skyturnred View Post
I apologize for being rude. Although it's no excuse, I have just been getting increasingly frustrated at simply not understanding what many others in my faculty seem to be able to understand in their sleep.

Also, I went to a terrible high school. Almost none of my graduating class went on to post-secondary education. I'm almost positive we didn't do series. I asked my friend who also went to my high school and he says the same thing.



Thanks for letting me know this, I wasn't quite understanding exactly what divergence meant. I thought a sum HAD to go off to infinity for it to be "divergent," but I guess it is considered "divergent" if it doesn't converge as well.
Re your statement ".. I have just been getting increasingly frustrated ..". You probably won't want to hear this, but: the way to gain proficiency is to do it again and again and again ... It's the old cliche that 'practice makes perfect'. Also: don't feel bad about making mistakes; just try to learn from them (although, of course, it would be nicer to not make mistakes on marked assignments).

Also: about having a single answer field: are you doing an on-line course, or are the assignments given on-line? Unfortunately, in this Forum it happens over and over again that people submit correct solutions to on-line questions and are marked wrong. There are some genuine unsolved technical issues with designing robust on-line assigment marking schemes.

RGV


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