Set of convergence of a Power series

[DottZakapa]

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?

 

[scottdave]

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?

 

[Mark44]

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.

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.

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.

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.

 

[DottZakapa]

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

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

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

 

[Mark44]

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

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

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

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.