Series and limit help for analysis

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Homework Help Overview

The discussion revolves around convergence of series and limits in the context of real analysis. Participants are exploring the implications of convergence for sequences and functions, particularly focusing on proving relationships between convergent series and limits of functions as they approach specific points.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the convergence of series, specifically questioning how the convergence of a series implies the convergence of its square. There are inquiries about the application of the comparison test and the conditions under which it can be used.
  • Questions arise regarding the existence of limits for specific functions, with participants expressing uncertainty about how to formally articulate why certain limits do not exist.
  • There is a focus on understanding the behavior of a piecewise function and establishing the uniqueness of limits based on the density of rational and irrational numbers.

Discussion Status

The discussion is ongoing, with participants sharing insights and questioning each other's reasoning. Some guidance has been offered regarding the application of the comparison test and the conditions for limits to exist, but there is no clear consensus on the proofs or concepts being discussed.

Contextual Notes

Participants express confusion about specific mathematical concepts and the formal definitions required to prove or disprove the convergence of series and the existence of limits. There is mention of constraints related to the participants' current understanding of continuity and limits.

JasonJo
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I need help proving that if (xn) is a convergent series, then (xn)^2 is a convergent series.

also is (n^3)*(xn) is convergent, then (xn) is convergent. i reasoned that if (n^3)*(xn) is convergent, we know that (n^3)*(xn) > (xn) > 0, therefore by comparison test, (xn) is convergent.

determine whether or not the function exists and if it doesn't explain why:

lim x--> 0 (x^2/(|x|))
i can't put it in formal terms why this limit does not exist (we did not go over continiuty yet)

show that the function
f(x) = { x if x is rational, 0 if x is irrational
has a limit at p=0 and does not have a limit at any other point.

can't quite establish that the limit exists and how do i prove that no other points have limits?
 
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JasonJo said:
I need help proving that if (xn) is a convergent series, then (xn)^2 is a convergent series.

You can use the comparison test here.

also is (n^3)*(xn) is convergent, then (xn) is convergent. i reasoned that if (n^3)*(xn) is convergent, we know that (n^3)*(xn) > (xn) > 0, therefore by comparison test, (xn) is convergent.

Looks good.

determine whether or not the function exists and if it doesn't explain why:

lim x--> 0 (x^2/(|x|))
i can't put it in formal terms why this limit does not exist (we did not go over continiuty yet)

What makes you think it doesn't exist? What does the function look like?

show that the function
f(x) = { x if x is rational, 0 if x is irrational
has a limit at p=0 and does not have a limit at any other point.

can't quite establish that the limit exists and how do i prove that no other points have limits?

For a limit to exist, it must be the same no matter how you approach the point. So if you can find two sequences x_n and y_n with both x_n->x and y_n->x, but f(x_n)->a and f(y_n)->b, then the limit at x doesn't exist.
 
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how could i use comparison test for the first one? sum(xn) is convergent, but sum(xn)^2 is greater than sum(xn) and therefore comparison test is not applicable, you need the greater series to be convergent to deduce that the lesser series is convergent
 
bump any help??

i still do not get how to solve the sum(an) converges then sum(an)^2 converges, prove or disprove

i still don't get how lim x--> 0 (x^2/(|x|)) has a limit or doesn't have a limit

or how to show that f(x) = {x if x is rational, 0 if x is irrational) has only a limit at p = 0 and there are not limits at any other point.

big help is needed, I am so lost
 
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JasonJo said:
how could i use comparison test for the first one? sum(xn) is convergent, but sum(xn)^2 is greater than sum(xn) and therefore comparison test is not applicable, you need the greater series to be convergent to deduce that the lesser series is convergent

(a_n)²is not necessarily greater than a_n. Take a_n=½ for instance. Then (a_n)²=¼<½. And it just so happen that when a series converges, the limit of its "argument" is 0, so for sufficiently large n, a_n<0 ==>a_n^2<a_n. So by the comparison test, (a_n)² converges.

You can also do it by the ratio tets. The esssence of the ratio test is that given two series [itex]\sum a_n[/itex], [itex]\sum b_n[/itex] for which you know that [itex]\sum a_n[/itex] converges, consider the limit of the ratio of their general term:

[tex]\frac{b_n}{a_n}[/tex]

and you know that since [itex]\sum a_n[/itex] converges, then [itex]a_n\rightarrow 0[/itex], so if the limit of the ratio is 0, then it means that b_n goes to 0 faster than a_n. Then surely, [itex]\sum b_n[/itex] will converge (more rigorously, it is so by the comparison test for sequences applies to the sequences of partial sums). Also, if the ratio "stabilizes" at infinity, i.e. equals some constant L, then b_n dives as fast as a_n so [itex]\sum b_n[/itex] must converge as well.

I don't know why I'm telling you this; it's for me I guess. But the point is that it is obvious that by taking a_n as your series and b_n = (a_n)², then it works.
 
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JasonJo said:
i still don't get how lim x--> 0 (x^2/(|x|)) has a limit or doesn't have a limit

Would it help you see it if you wrote x² as |x|²?

JasonJo said:
or how to show that f(x) = {x if x is rational, 0 if x is irrational) has only a limit at p = 0 and there are not limits at any other point.

What do you know about the density of the rationals and inrationals is R?

Use that and the fact that if the limit exists, it is unique.
 

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