- #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?

##\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?