# Can I Construct an Infinite Series that Approaches a Specific Limit?

• thetexan
In summary, the conversation discusses finding the limit of a series in reverse, specifically whether there is a way to come up with an infinite series that approaches a pre-chosen limit. The approach for this involves using an implicit function algorithm to find the roots of the equation f(x) = 0, taking into account the continuity and convergence of the series. Examples of infinite series that converge to specific values are also given. The questioner clarifies that they are interested in more general series, such as the harmonic series, rather than simple sequences.
thetexan
I know you can determine what number the limit of a series approaches, such as 2. Is there a way to do that in reverse? Is there a method where I can I come up with an infinite series that approaches a limit of, say, 7.5? or 11.75? And is the possible different series that will approach, say 5.25, infinite or finite?

tex

Have you tried to come up with infinite series with a pre-chosen sum? It's a rather trivial exercise... unless you're have a mental block that's preventing you from doing easy things.

thetexan said:
I know you can determine what number the limit of a series approaches, such as 2. Is there a way to do that in reverse? Is there a method where I can I come up with an infinite series that approaches a limit of, say, 7.5? or 11.75? And is the possible different series that will approach, say 5.25, infinite or finite?

tex

Hey thetexan.

This is pretty much the same as finding the inverse of a function value. You need to aware of a few things:

There must be a one-to-one relationship between the value you have and the value of x ot t or whatever in the series expansion. The limit must also exist in some neighbourhood and you want to have continuity in an appropriate neighbourhood for when you have to use a numerical routine which will have errors associated with it when you are dealing with a general expression (limited numbers of expressions can be solved analytically).

If you have these (and maybe other conditions), then if you can assume continuity and convergence in the right neighbourhood then you can use an implicit function algorithm that finds the roots to the equation f(x) = 0 (I'm assuming one dimension is your series, if it's two the same ideas apply but it's more complex and you need more conditions met).

Then when you find the solutions, you need to take into account the nature of the series itself. If the series is a standard analytic series, then the continuity and convergence should be known before hand and if you get these two properties across the whole domain then you don't have to worry.

The point where you have to worry is where you don't get this continuity or convergence set of conditions because you can get limits for things that are not analytic like say the function |x| = absolute value of x.

If you can write your series as some kind of power series and you can gaurantee convergence in the appropriate region, then just use the implicit function solver: it's also known as a root finding algorithm and many popular math packages have this as a library or as a simple function call.

Well, no. Thus the question. If it is a trivial exercise could you please show me how?

Hurkyl said:
Have you tried to come up with infinite series with a pre-chosen sum? It's a rather trivial exercise... unless you're have a mental block that's preventing you from doing easy things.

I don't think this is true. Consider the relationship sin(x) = a for a general a. It's not going to be easy to find x for all possible a in the principal branch. This is just one candidate of many in the context of the OP's question.

thetexan said:
Well, no. Thus the question. If it is a trivial exercise could you please show me how?

Are you talking about a general power series? So basically an 'infinite-polynomial' type power series that is gauranteed to converge and has some known inverse region?

chiro said:
I don't think this is true. Consider the relationship sin(x) = a for a general a. It's not going to be easy to find x for all possible a in the principal branch. This is just one candidate of many in the context of the OP's question.
What does anything you've said have to do with the question? I think you're having a mental block too.Here are three easy examples:

$$7.5 + 0 + 0 + 0 + 0 + \cdots = 7.5$$
$$4 + 3.5 + 0 + 0 + \cdots = 7.5$$
$$7 + 1/4 + 1/8 + 1/16 + 1/32 + \cdots = 7.5$$​

Hurkyl said:
What does anything you've said have to do with the question? I think you're having a mental block too.Here are three easy examples:

$$7.5 + 0 + 0 + 0 + 0 + \cdots = 7.5$$
$$4 + 3.5 + 0 + 0 + \cdots = 7.5$$
$$7 + 1/4 + 1/8 + 1/16 + 1/32 + \cdots = 7.5$$​

Yes but that's not a general solution. I think the author wants to analyze a general form of an expression. With the exception of the last one, most of those are more representative of a sequence rather than a series don't you think?

Also I am referring to an arbitrary sequence that is not based on a formula as opposed to a sequence that is if you need any further clarification.

It's fairly easy to construct a geometric series that converges to any desired value (actually, you can construct an infinite number of them).

Number Nine said:
It's fairly easy to construct a geometric series that converges to any desired value (actually, you can construct an infinite number of them).

Before I comment further I would like to know from the OP exactly what kinds of series he is referring to, because if he only wants to consider a limit class of geometric series then you and Hurkyl have given a sufficient answer, but if it is more general then the answer will of course change.

First, I am only a neophyte math hobbiest with not a very good understanding of even this subject.

When I say series I think about things like the harmonic series, and other series that contain algebraic parts (meaning, to me, terms such as ((x^s+(1-x^3)) ( I just made that up)). Sure 7.5 + 0 + 0 + 0 + 0 is a series but that's not what I mean. I may not even know enough to ask an intelligent question. But when I read about primes and other math work I see these complex series where they describe that they limit to a number such as 2 or 1.5 or such. It occurred to me to ask, 'I wonder if you can do that in reverse?'.

My question is can some math method be used to discover a series that when added together to a limit produces a predetermined sum...let's say 13.564...? or 4.1? or .66? or 8?

tex

OK, let's say you want to make a geometric series converge to $x$ where we require $0 < x <$, and where we want a series in the form $\sum a^n$ where $|a| < 1$. Now, we know (or at least some of us know, I don't know if you have covered this) that
$$\sum a^n = \frac{1}{1-a}$$

OK, so let's try this:
$$\frac{1}{1-a} = \frac{1}{x}$$

And let's solve for a:
$$a = 1 - x$$

Since $|x| < 1$ we have that $|1-x| < 1$, so this is a valid $a$. Now we have this series:
$$\sum a^n = \sum (1-x)^n = \frac{1}{1-(1-x)} = \frac{1}{x}$$

Now, this isn't $x$. So, let's take $r = x^2$. Now, we know that
$$\sum ra^n = \frac{r}{1-a}$$

So, let's take $r = x^2$ so that we have:

$$\sum ra^n = \sum x^2(1-x)^n = \frac{x^2}{1-(1-x)} = x$$

So, unless I screwed up somewhere, this is a series that converges to and $x$ such that $0 < x < 1$. Now, do you see how you can take this method and adjust it by adding a term and/or multiplying the series by a constant that will cause it to converge to any given value?NOTE: I think most standard texts reverse the roles $r$ and $a$ from the way that I have done it. I always forget the formula, though, and end up having to re-derive it, so my apologies if this causes any confusion, but you should still be able to follow what I have written (at least, that transposition of $r$ and $a$ shouldn't keep you from understanding).

Last edited:
thetexan said:
First, I am only a neophyte math hobbiest with not a very good understanding of even this subject.

When I say series I think about things like the harmonic series, and other series that contain algebraic parts (meaning, to me, terms such as ((x^s+(1-x^3)) ( I just made that up)). Sure 7.5 + 0 + 0 + 0 + 0 is a series but that's not what I mean. I may not even know enough to ask an intelligent question. But when I read about primes and other math work I see these complex series where they describe that they limit to a number such as 2 or 1.5 or such. It occurred to me to ask, 'I wonder if you can do that in reverse?'.

My question is can some math method be used to discover a series that when added together to a limit produces a predetermined sum...let's say 13.564...? or 4.1? or .66? or 8?

tex

In this case if you are looking at a general power series, then what you can do is pick any series that converges and is continuous around a neighbourhood of the point you have, and then after that it can get quite messy.

You could use the properties of convergence for the series and use a root finding algorithm (provided the conditions are right) to get an approximation for your x, or t or whatever: it will not be exact of course but if you provide a decimal limit with some resolution and your answer for x or t gives an answer that has an error less than the resolution then you are done (you say that there is an error term though).

Other ways that I can think of is to use relationships between L^2 and l^2 using any kind of integral transform. The Basel problem can be solved this way by considering the function y = x in the region of [-pi,pi] and then using norm identities to show equivalence. Here is the wiki page:

http://en.wikipedia.org/wiki/Basel_problem

The tricky part for the above though is that you need to figure out what the function should be for integrating and this is far from easy.

Robert1986 said:
OK, let's say you want to make a geometric series converge to $x$ where we require $0 < x <$, and where we want a series in the form $\sum a^n$ where $|a| < 1$. Now, we know (or at least some of us know, I don't know if you have covered this) that
$$\sum a^n = \frac{1}{1-a}$$

OK, so let's try this:
$$\frac{1}{1-a} = \frac{1}{x}$$

And let's solve for a:
$$a = 1 - x$$

Since $|x| < 1$ we have that $|1-x| < 1$, so this is a valid $a$. Now we have this series:
$$\sum a^n = \sum (1-x)^n = \frac{1}{1-(1-x)} = \frac{1}{x}$$

Now, this isn't $x$. So, let's take $r = x^2$. Now, we know that
$$\sum ra^n = \frac{r}{1-a}$$

So, let's take $r = x^2$ so that we have:

$$\sum ra^n = \sum x^2(1-x)^n = \frac{x^2}{1-(1-x)} = x$$

So, unless I screwed up somewhere, this is a series that converges to and $x$ such that $0 < x < 1$. Now, do you see how you can take this method and adjust it by adding a term and/or multiplying the series by a constant that will cause it to converge to any given value?

NOTE: I think most standard texts reverse the roles $r$ and $a$ from the way that I have done it. I always forget the formula, though, and end up having to re-derive it, so my apologies if this causes any confusion, but you should still be able to follow what I have written (at least, that transposition of $r$ and $a$ shouldn't keep you from understanding).

I realize I was a bit unclear when I mentioned using this method to create a more general sequence. If $|x| > 1$ just note that $|1/x| < 1$ to make a power series in the more general sense.

chiro said:
In this case if you are looking at a general power series, then what you can do is pick any series that converges and is continuous around a neighbourhood of the point you have, and then after that it can get quite messy.

You could use the properties of convergence for the series and use a root finding algorithm (provided the conditions are right) to get an approximation for your x, or t or whatever: it will not be exact of course but if you provide a decimal limit with some resolution and your answer for x or t gives an answer that has an error less than the resolution then you are done (you say that there is an error term though).

Other ways that I can think of is to use relationships between L^2 and l^2 using any kind of integral transform. The Basel problem can be solved this way by considering the function y = x in the region of [-pi,pi] and then using norm identities to show equivalence. Here is the wiki page:

http://en.wikipedia.org/wiki/Basel_problem

The tricky part for the above though is that you need to figure out what the function should be for integrating and this is far from easy.

Unless I'm just way out of the neighbourhood (pun intended), I think the way I described will do what he wants, won't it?

Robert1986 said:
Unless I'm just way out of the neighbourhood (pun intended), I think the way I described will do what he wants, won't it?

It depends on how general he wants to go with the series definition.

Your series expansion has the same coeffecient for each term of a given power in the series. Since the OP mentioned things like harmonic series, it made sense for me to interpret more general forms of series. In that vein, I mentioned what I mentioned above.

Your example is basically for constructing a very limited class of series and if the OP wants to know what to do for more complex series like a harmonic series (which he mentioned above) then you can't just resort only to examples like that.

I don't think the OP realizes how difficult it is to expand not only power series with arbitrary coeffecients, but even the general series like say the Riemann Zeta function (I brought this up because he did mention an interest in number theory).

chiro said:
It depends on how general he wants to go with the series definition.

Your series expansion has the same coeffecient for each term of a given power in the series. Since the OP mentioned things like harmonic series, it made sense for me to interpret more general forms of series. In that vein, I mentioned what I mentioned above.

Your example is basically for constructing a very limited class of series and if the OP wants to know what to do for more complex series like a harmonic series (which he mentioned above) then you can't just resort only to examples like that.

I don't think the OP realizes how difficult it is to expand not only power series with arbitrary coeffecients, but even the general series like say the Riemann Zeta function (I brought this up because he did mention an interest in number theory).

Ahhh, yes, very true.

## 1. What is a limit in mathematics?

A limit in mathematics is the value that a function or sequence approaches as the input or index approaches a certain value. It is used to describe the behavior of a function or sequence around a specific point or as the input or index approaches infinity or negative infinity.

## 2. How do you calculate a limit?

To calculate a limit, you must first identify the function or sequence and the value that the input or index is approaching. Then, use algebraic techniques, such as factoring or simplifying, to manipulate the function or sequence into a form that can be evaluated. Finally, substitute the approaching value into the simplified function or sequence to find the limit.

## 3. What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit only considers the behavior of a function or sequence as the input or index approaches from one direction, either from the positive or negative side. A two-sided limit takes into account the behavior of the function or sequence from both directions, approaching from the positive and negative side. One-sided limits are used when the function or sequence has a discontinuity or a vertical asymptote, while two-sided limits are used when the function or sequence is continuous.

## 4. What is the significance of limits in calculus?

Limits are fundamental to the study of calculus as they are used to define the concepts of continuity, derivatives, and integrals. They allow us to understand the behavior of a function or sequence at a specific point or as the input or index approaches infinity, which is essential in understanding the rate of change and accumulation of quantities in calculus.

## 5. Can limits be used to prove the existence of a limit?

Yes, limits can be used to prove the existence of a limit. To prove the existence of a limit, we must show that the function or sequence approaches the same value regardless of the approaching value. This can be done by using the definition of a limit or by using various limit theorems, such as the Squeeze theorem or the Intermediate Value theorem.

Replies
7
Views
2K
Replies
1
Views
1K
Replies
11
Views
1K
Replies
4
Views
1K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
15
Views
3K
Replies
4
Views
2K
Replies
2
Views
979
Replies
3
Views
1K