Set of convergence of a Power series

In summary, a power series is an infinite series used to represent a function as a sum of infinitely many terms. The set of convergence of a power series is the set of all values of x for which the series converges, and it can be determined using the ratio or root test. The radius of convergence is the distance from the center of the series to the nearest point where the series converges, and the interval of convergence is given by (a-R, a+R), where a is the center and R is the radius. The endpoints of the interval may or may not be included in the set of convergence.
  • #1
DottZakapa
239
17
Homework Statement
Find convergence radius and set of convergence,
Relevant Equations
power series
given the following

##\sum_{n=0}^\infty n^2 x^n##

in order to find the radius of convergence i do as follows

##\lim_{n \rightarrow +\infty} \left |\sqrt [n]{n^2}\right|=1##

hence the radius of convergence is R=##\frac 1 1=1##

|x|<1

Now i have to verify how the series behaves at the extrema points of the convergence set.

for x=-1

##\sum_{n=0}^\infty n^2 (-1)^n##

at this point i should solve

##\lim_{n \rightarrow +\infty} {n^2} (-1)^n##

cannot apply leibniz because the ##\lim_{n \rightarrow +\infty} {n^2}## doesn't go to zero

I did not really understand how to handle the (-1)^n, normally such limit does not exists, is this the right conclusion?

is it correct in certain other cases to consider leibniz if possible(according to the conditions of the theorem) or the absolute value of ##a_n## an then use the tests in order to check the convergence or divergence at such points?
 
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  • #2
So you have some series where the sequence converges to zero, but the series (sum) diverges, like a Harmonic Series. An alternating harmonic sequence converges to zero as well. What can you say about your alternating sequence?
 
  • #3
DottZakapa said:
Homework Statement:: Find convergence radius and set of convergence,
Relevant Equations:: power series

given the following

##\sum_{n=0}^\infty n^2 x^n##

in order to find the radius of convergence i do as follows

##\lim_{n \rightarrow +\infty} \left |\sqrt [n]{n^2}\right|=1##
This is a power series, so |x| should appear in the equation above.
DottZakapa said:
hence the radius of convergence is R=##\frac 1 1=1##

|x|<1

Now i have to verify how the series behaves at the extrema points of the convergence set.

for x=-1

##\sum_{n=0}^\infty n^2 (-1)^n##

at this point i should solve

##\lim_{n \rightarrow +\infty} {n^2} (-1)^n##

cannot apply leibniz because the ##\lim_{n \rightarrow +\infty} {n^2}## doesn't go to zero
Your textbook should list the N-th Term Test for Divergence, which says that if the n-th term of a series isn't zero or doesn't exist, the series diverges.
DottZakapa said:
I did not really understand how to handle the (-1)^n, normally such limit does not exists, is this the right conclusion?
No. The presence of a factor of ##(-1)^n## usually indicates an alternating series, which could converge or diverge, depending on the other factors.
DottZakapa said:
is it correct in certain other cases to consider leibniz if possible(according to the conditions of the theorem) or the absolute value of ##a_n## an then use the tests in order to check the convergence or divergence at such points?
If the series is alternating, then Leibniz Test applies. This test is also called the Alternating Series Test.
 
  • #4
i have an exercise solved in class on my notes done as follows
##\sum_{n=1}^\infty n^2 (-1)^n##

##\nexists~\lim_{n \rightarrow +\infty} {n^2} (-1)^n~\Rightarrow## the necessary condition of convergence is not satisfied hence it is not convergent

same here

##\sum_{n=1}^\infty n (-1)^n##

##\nexists~\lim_{n \rightarrow +\infty} {n} (-1)^n##

:oldconfused:
 
  • #5
DottZakapa said:
i have an exercise solved in class on my notes done as follows
##\sum_{n=1}^\infty n^2 (-1)^n##

##\nexists~\lim_{n \rightarrow +\infty} {n^2} (-1)^n~\Rightarrow## the necessary condition of convergence is not satisfied hence it is not convergent

same here

##\sum_{n=1}^\infty n (-1)^n##

##\nexists~\lim_{n \rightarrow +\infty} {n} (-1)^n##

:oldconfused:
Since neither limit exists, neither series converges. This is the N-th Term Test for Divergence in operation, that I mentioned earlier. See https://en.wikipedia.org/wiki/Term_test.
 
  • Informative
Likes scottdave

Related to Set of convergence of a Power series

1. What is a power series?

A power series is an infinite series of the form n=0∞ an(x−c)n, where an and c are constants and x is a variable. It is a type of infinite polynomial that can be used to represent certain functions.

2. What does it mean for a power series to converge?

A power series is said to converge if its terms approach a finite value as the number of terms increases. In other words, the sum of the series approaches a finite value as more and more terms are added. If the sum does not approach a finite value, the series is said to diverge.

3. How do you determine the set of convergence for a power series?

The set of convergence for a power series is determined by finding the values of x for which the series converges. This can be done using various convergence tests, such as the ratio test or the root test. The set of convergence will typically be an interval centered at the point c in the series.

4. What is the radius of convergence for a power series?

The radius of convergence for a power series is the distance from the center point c to the nearest point where the series converges. It can be determined using the ratio test or the root test, and is often denoted by R.

5. Can a power series converge at its endpoints?

It is possible for a power series to converge at one or both of its endpoints, but this is not always the case. The convergence at the endpoints depends on the behavior of the series at these points and cannot be determined solely based on the convergence at other points within the interval of convergence.

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