# Series and limit help for analysis

• JasonJo
In summary, if (xn) is a convergent series, then (xn)^2 is also a convergent series. The comparison test and the ratio test can be used to prove this. Additionally, if (n^3)*(xn) is convergent, then (xn) is also convergent. It can be reasoned that if (n^3)*(xn) is convergent, then (n^3)*(xn) > (xn) > 0, and by comparison test, (xn) is convergent. For the function f(x) = { x if x is rational, 0 if x is irrational }, it has a limit at p=0 and does not have a limit at any other point. This can be
JasonJo
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?

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 $\sum a_n$, $\sum b_n$ for which you know that $\sum a_n$ converges, consider the limit of the ratio of their general term:

$$\frac{b_n}{a_n}$$

and you know that since $\sum a_n$ converges, then $a_n\rightarrow 0$, so if the limit of the ratio is 0, then it means that b_n goes to 0 faster than a_n. Then surely, $\sum b_n$ 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 $\sum b_n$ 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.

## 1. What is a series in analysis?

A series in analysis is a sequence of numbers that are added together to obtain a sum. Each term in the series is referred to as a "partial sum".

## 2. What is the difference between a finite and infinite series?

A finite series has a limited number of terms, while an infinite series has an infinite number of terms. In other words, a finite series has an ending point, while an infinite series does not.

## 3. What is a limit in analysis?

A limit is a fundamental concept in calculus and analysis that describes the behavior of a function as its input approaches a certain value. It is represented by the notation lim f(x) as x approaches a.

## 4. How is the limit of a series calculated?

The limit of a series can be calculated using various methods, such as the ratio test, root test, or comparison test. These tests compare the series to known series with known convergence behaviors to determine the convergence or divergence of the given series.

## 5. What is the importance of series and limits in analysis?

Series and limits are important in analysis as they allow us to study the behavior of functions and sequences in a specific context. They are used to determine the convergence or divergence of a series, which is crucial in many mathematical and scientific applications.

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