Power Series problem

[Bliman]

Hi all,
I have this power series problem and the convergence of it that I don't understand is the following one the summation of
(k(k+1)(k+2)***(k+n-1)x^n)/n! from n=0 to infinity
Now you solve this with the ratio test and you do this with lim as n approaches infinity of the absolute value of (an+1)/an and solve it for less than one.
Now what I don't get is why you don't do (k+(n+1)-1) but instead skip over it and do this (k+n-1)(k+n). Why not do the (n+1) directly instead of leaving (k+n-1) alone and multiplying it then with (k+n)?
I hope it is clear because I don't find how you put mathematical symbols in the question (the read before posting didn't really help).
Also these *** always confuses me, sometimes you have to put them in factorials but that is super confusing for me. Where can I learn more how to handle this?
Thanks for helping me

 

[LCKurtz]

It isn't clear to me exactly what it bothering you. Did you get this:
[MATH]\left| \frac {a_{n+1}}{a_n}\right | = \frac {k+n}{n+1} |x|[/MATH]when you did the ratio? Or is the cancellation getting to that what's bothering you?

 

[Bliman]

Hi,
It is this question . You see that in the original question it is stated (k+n-1)x^n. But I have looked to professor Leonard and such (I am self-learning) and also following the Larson Calculus book. That with the ratio test every time you see an n you change it to an (n+1) and then divide by the same function but just like it is.
But as you can see they skip over the first because they leave the (k+n-1) alone and then create an extra (k+n). And I don't know why don't they just have a (k+n). Why the extra (k+n-1)?
I also have problems in general with those 3 points. Beneath there is also one I struggle with . Again with the 3 points because I don't know how to put them in the good factorial. And I would like to know how I can do better on that or see what to do with these problems easier. So those are basically two different questions.
Thanks for helping by the way.

 

[Steven G]

I am looking at the image above number 39.

In the very first line it has k(k+1)...(k+n-1). Are you wondering why it does not have k(k+1)....(k+n)? If yes, then you have to understand that you can't change the given problem. The ratio test does NOT suggest that you change the original problem. Rather it says, with a proof, that you compute the limit of a_n+1/a_n. Now aⁿ⁺¹*1/a_n is what it says on the 2nd line simplified.

You do realize that k+(n+1)-1 = k+n

Please state exactly where you are having trouble.

For a proof of the ratio test go here

 

[pka]

I am looking at the image above number 39.

Indeed, I too have questions about #39. What is \(k\) doing in that series? \(k\) is nowhere defined, it is just there with no explanation.

 

[Steven G]

I assumed that k is some real number.

 

[Bliman]

Yeah sorry,
k is greater or equal than one.
Yes the first line has (k+n-1). But by the ratio test, I thought that you had to replace the n with (n+1).
That way you have like you said (k+(n+1)-1)=(k+n). So you would have in the ratio test k(k+1)...(k+n)x^n But as you can see in the second line where they do the ratio test there is not only (k+n) but before it also the same (k+n-1).
Is this clear enough?

 

[Dr.Peterson]

Hi,
It is this question . You see that in the original question it is stated (k+n-1)x^n. But I have looked to professor Leonard and such (I am self-learning) and also following the Larson Calculus book. That with the ratio test every time you see an n you change it to an (n+1) and then divide by the same function but just like it is.
But as you can see they skip over the first because they leave the (k+n-1) alone and then create an extra (k+n). And I don't know why don't they just have a (k+n). Why the extra (k+n-1)?
I also have problems in general with those 3 points. Beneath there is also one I struggle with . Again with the 3 points because I don't know how to put them in the good factorial. And I would like to know how I can do better on that or see what to do with these problems easier. So those are basically two different questions.
Thanks for helping by the way.

Your question is about how they go from [MATH]k(k+1)\cdots(k+n-1)[/MATH] to [MATH]k(k+1)\cdots(k+n-1)(k+n)[/MATH], right?

As you said, they are replacing [MATH]n[/MATH] with [MATH]n+1[/MATH]. When you replace [MATH]n[/MATH] with [MATH]n+1[/MATH] in [MATH](k+n-1)[/MATH], you get [MATH](k+(n+1)-1) = (k+n)[/MATH]. That's the new last factor.

And how about the factor before that? In the original, that would be [MATH](k+n-2)[/MATH], right? When you replace [MATH]n[/MATH] with [MATH]n+1[/MATH], that becomes [MATH](k+(n+1)-2) = (k+n-1)[/MATH]. They write that explicitly to make it easier to see the next step, where they have to cancel [MATH]k(k+1)\cdots(k+n-1)[/MATH].

But you can also think of it this way: the original is the product of all integers (if [MATH]k[/MATH] is an integer) from [MATH]k[/MATH] until the one before [MATH]k+n[/MATH]. Changing [MATH]n[/MATH] to [MATH]n+1[/MATH] means that you want all integers from [MATH]k[/MATH] through [MATH]k+n[/MATH] -- they are indeed adding on one more factor, because that's what the expression means. Changing [MATH]n[/MATH] to [MATH]n+1[/MATH] means going one step further.

Another way to help yourself see what is happening is to pick some sample numbers and check out what it means.

Suppose [MATH]k=5[/MATH] and [MATH]n=4[/MATH]. Then [MATH]k(k+1)\cdots(k+n-1)[/MATH] is [MATH]5(5+1)\cdots(5+4-1) = (5)(6)\cdots(8)[/MATH], that is [MATH](5)(6)(7)(8)[/MATH].

Now when we increase [MATH]n[/MATH] to [MATH]5[/MATH], we have [MATH]5(5+1)\cdots(5+5-1) = (5)(6)\cdots(9)[/MATH], that is [MATH](5)(6)(7)(8)(9)[/MATH]. Do you see that we have added on one more factor?

In #31, they left out the product form entirely, skipping to after the cancellation. See if you can follow it now, by writing in the skipped steps.

 

[Singleton]

k(k+1)...(k+n-1)(k+n) is the same as k(k+1)...(k+n) it's just writing out one more term (k+n-1) that's in the "..." right before (k+n).
Reason for doing this is now you can see part of this product matching the same thing in the denominator. Then you cancel.
This is something you will see again and often in these calculations. So it's good you asked!

 

[Bliman]

Thank you all for responding.
Dr.Peterson hit the nail on the head. Thank you for the great explanation.
I didn't know precisely what those three points stood for. And I am not exactly clear now either.
Take for example Suppose k=5 and n=4. Then k(k+1)⋯(k+n−1) is 5(5+1)⋯(5+4−1)=(5)(6)⋯(8), that is (5)(6)(7)(8).
Imagine in another function you have (5)(7)⋯(13). How do you intrepet the points then. Is it then (5)(7)(8)(9)(10)(11)(12)(13) or rather (5)(7)(9)(11)(13)?
Again with the second example (31) you have (2)(3)(4)⋯(n+1).
If you take n=1 you have (2)(3)(4)⋯(2). Would you have then (2)(3)(4)(3)(2)? So you would get (n+1)(n+2)(n+3)(n+2)(n+1)? Is that correct?
Thanks for helping me. I have always been to shy to come in such fora as these. So many great minds and me as probably a stupid person.

 

[tkhunny]

I have always been to shy to come in such fora as these.

Delighted to see you overcoming this.

... and me as probably a stupid person.

Perhaps it is time to overcome this? Not much chance of success if that's how you think about yourself. It's fine to have questions. It's very helpful, as has been demonstrate, to ask those questions.

You posted your question.
You showed your work.
You interacted with the other participants.
You continued to question until you began to understand.
This has NOTHING to do with "stupid person".

Glad to have you with us. We're all happy to help when you act like that.

 

[Dr.Peterson]

Thank you all for responding.
Dr.Peterson hit the nail on the head. Thank you for the great explanation.
I didn't know precisely what those three points stood for. And I am not exactly clear now either.
Take for example Suppose k=5 and n=4. Then k(k+1)⋯(k+n−1) is 5(5+1)⋯(5+4−1)=(5)(6)⋯(8), that is (5)(6)(7)(8).
Imagine in another function you have (5)(7)⋯(13). How do you intrepet the points then. Is it then (5)(7)(8)(9)(10)(11)(12)(13) or rather (5)(7)(9)(11)(13)?
Again with the second example (31) you have (2)(3)(4)⋯(n+1).
If you take n=1 you have (2)(3)(4)⋯(2). Would you have then (2)(3)(4)(3)(2)? So you would get (n+1)(n+2)(n+3)(n+2)(n+1)? Is that correct?

I didn't know what you meant by "those three points". It now occurs to me that you mean the three dots, k(k+1)⋯(k+n−1) .

They are simultaneously multiplications and an ellipsis, "...". They mean, "continue multiplying in the same pattern". The pattern involved is not explicit, but is understood to be successive numbers, as suggested by the first two which differ by 1. We see this often enough that we understand it as meaning that, and I would think you would have been told that at some point; but I can understand not seeing it at first.

So for example, [MATH]6\cdot7\cdots12[/MATH] would mean [MATH]6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12[/MATH].

I don't know how often you would see such expressions skipping more than one, but I would take [MATH]5\cdot7\cdots13[/MATH] to mean [MATH]5\cdot7\cdot9\cdot11\cdot13[/MATH], continuing by 2's. Any more complicated pattern would have to be stated explicitly.

There are times, as in your last question, when you can't take it literally. [MATH]2\cdot3\cdot4\cdots2[/MATH] would just mean [MATH]2[/MATH]: you are counting by 1's "from 2 to 2", which means you have only that one factor. You'll often see this in definitions of the factorial function, where [MATH]k\cdot(k-1)\cdot(k-2)\cdots1[/MATH] doesn't mean that there have to be at least four factors, so [MATH]k[/MATH] can't be less than 4!

 

[Bliman]

Delighted to see you overcoming this.



Perhaps it is time to overcome this? Not much chance of success if that's how you think about yourself. It's fine to have questions. It's very helpful, as has been demonstrate, to ask those questions.

You posted your question.
You showed your work.
You interacted with the other participants.
You continued to question until you began to understand.
This has NOTHING to do with "stupid person".

Glad to have you with us. We're all happy to help when you act like that.

Thank you for your kind words.
When I was in high school in the first and second grades (Belgium) (I followed biological science) I couldn't follow anything. I was totally overwhelmed by everything and I got horrible exam results. I had to choose another direction and I had very limited choices. So I choose to study as a cook. I always loved and still do anything related to science, philosophy, and math. Now I am 43 years old and it still wreaks havoc in my brain. I have been diagnosed with Asperger syndrome a few years ago. And I am now trying to learn many things on myself to see if it was Aspergers that was bothering me then or if I am just not smart enough. So thank you for the message. It helps more then you think.

 

[Bliman]

I didn't know what you meant by "those three points". It now occurs to me that you mean the three dots, k(k+1)⋯(k+n−1) .

They are simultaneously multiplications and an ellipsis, "...". They mean, "continue multiplying in the same pattern". The pattern involved is not explicit, but is understood to be successive numbers, as suggested by the first two which differ by 1. We see this often enough that we understand it as meaning that, and I would think you would have been told that at some point; but I can understand not seeing it at first.

So for example, [MATH]6\cdot7\cdots12[/MATH] would mean [MATH]6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12[/MATH].

I don't know how often you would see such expressions skipping more than one, but I would take [MATH]5\cdot7\cdots13[/MATH] to mean [MATH]5\cdot7\cdot9\cdot11\cdot13[/MATH], continuing by 2's. Any more complicated pattern would have to be stated explicitly.

There are times, as in your last question, when you can't take it literally. [MATH]2\cdot3\cdot4\cdots2[/MATH] would just mean [MATH]2[/MATH]: you are counting by 1's "from 2 to 2", which means you have only that one factor. You'll often see this in definitions of the factorial function, where [MATH]k\cdot(k-1)\cdot(k-2)\cdots1[/MATH] doesn't mean that there have to be at least four factors, so [MATH]k[/MATH] can't be less than 4!

Ok thank you for the help.
Yeah, those ellipsis always put me in a bind.
So if I am correct if there is an obvious pattern then you should go for the pattern (with explicit)?
So if in example 31 n is equal or greater then 4 then you have n factorial but if you have n equals 1 then you have (2)(3)(4)(2) because there is no pattern established.
I hope you are not tearing your hair out. And I hope my above assessment is the right one.

 

[Dr.Peterson]

So if I am correct if there is an obvious pattern then you should go for the pattern (with explicit)?
So if in example 31 n is equal or greater then 4 then you have n factorial but if you have n equals 1 then you have (2)(3)(4)(2) because there is no pattern established.

No, in example 31, [MATH]2\cdot3\cdot4\cdots(n+1)[/MATH] is always equal to [MATH](n+1)![/MATH]. It is not different when [MATH]n<4[/MATH]. In the latter case, it is not to be taken literally as having to start with [MATH]2\cdot3\cdot4[/MATH].

You can think of this as something like an idiom in English, which is something we say that is not to be taken literally. We say that "everyone knows" what we mean; if you are not familiar with an idiom, you have to learn it, rather than fight against it. (My son used to complain when he was called "sweetie pie", "I am not a pie!". He was literally right, but this was cute, funny, and childish, and if he seriously thought that was what was meant, we would worry.)

So when we read [MATH]2\cdot3\cdot4\cdots(n+1)[/MATH], we read that as, "Starting with 2 and increasing by 1 each time, write numbers ending with (n+1), and multiply these together. The 3 and 4 are taken as an illustration of the pattern, not as a necessary part of what the expression represents. We identify the pattern, and then we back off and apply it even to cases that don't literally match what is written.

 

[Bliman]

No, in example 31, [MATH]2\cdot3\cdot4\cdots(n+1)[/MATH] is always equal to [MATH](n+1)![/MATH]. It is not different when [MATH]n<4[/MATH]. In the latter case, it is not to be taken literally as having to start with [MATH]2\cdot3\cdot4[/MATH].

You can think of this as something like an idiom in English, which is something we say that is not to be taken literally. We say that "everyone knows" what we mean; if you are not familiar with an idiom, you have to learn it, rather than fight against it. (My son used to complain when he was called "sweetie pie", "I am not a pie!". He was literally right, but this was cute, funny, and childish, and if he seriously thought that was what was meant, we would worry.)

So when we read [MATH]2\cdot3\cdot4\cdots(n+1)[/MATH], we read that as, "Starting with 2 and increasing by 1 each time, write numbers ending with (n+1), and multiply these together. The 3 and 4 are taken as an illustration of the pattern, not as a necessary part of what the expression represents. We identify the pattern, and then we back off and apply it even to cases that don't literally match what is written.

Thank you very much for sticking with me.
I think (I hope) I got it. So in the 31 example the (2)(3)(4) are basically examples to show how the formula works but you don't take these numbers as part of the problem and (n+1) is the formula which you work with. And in the 39 example you get a formula first and a formula at the end and you work with both.
Great teaching by the way and thank you for the patience.

 

[Dr.Peterson]

Thanks. I think you've got it.

I found this that I wrote a decade ago about the similar issue of ellipsis in a number. As I mention there, it should only be used in simple cases where there's no doubt about the pattern. If you need to be clearer, you would use pi (product) notation, which works much like sigma notation for summations, letting you be explicit about each factor.

 

[Bliman]

Thanks. I think you've got it.

I found this that I wrote a decade ago about the similar issue of ellipsis in a number. As I mention there, it should only be used in simple cases where there's no doubt about the pattern. If you need to be clearer, you would use pi (product) notation, which works much like sigma notation for summations, letting you be explicit about each factor.

Thank you very much for helping me. I shall read the links too. Thanks for being there for people like me.
And that goes for all the people that are helping here.