There is a Serie $\sum_{n=1}^{\infty} \frac{x^{2n}}{n^{2}}$
Convergence radius $r = 1$ is convergent in both endpoints
Set $y=x^{2}$ Look for convergence radius $r$ of Serie $\sum_{n=1}^{\infty} \frac{y^{n}}
{n^{2}}$. Let $a_{n} = \frac{1}{n^{2}}$ ...
I cant understand why or how we can take $a_{n} = \frac{1}{n^{2}}$

Question

There is a Serie $\sum_{n=1}^{\infty} \frac{x^{2n}}{n^{2}}$
Convergence radius $r = 1$ is convergent in both endpoints
Set $y=x^{2}$ Look for convergence radius $r$ of Serie $\sum_{n=1}^{\infty} \frac{y^{n}}
{n^{2}}$. Let $a_{n} = \frac{1}{n^{2}}$ ...
I cant understand why or how we can take $a_{n} = \frac{1}{n^{2}}$

anonymous · Answer

also my problem is with $$a_{n} = \frac{1}{n^{2}}$$ why we can take it instead of$$\frac{y^{n}}{n^{2}}$$

zzr0ck3r · Answer

a_n = (1/n)^2

blockcolder · Answer

If you examine the definition of a power series, it says something about a series of the ff. form:
$$\sum c_nx^n$$
So in this case, you ignore the variable raised to n (which is y^n) and consider the remaining expression in n.

zzr0ck3r · Answer

I'm a little confused by the question

anonymous · Answer

hmm i am confusing in instead of $$y^{n}$$ why we write 1

anonymous · Answer

@blockcolder can you tell me please in other words how we can get $$1$$ instead of $$y^{n}$$

blockcolder · Answer

You don't replace $y^n$ by 1.
$$\frac{y^n}{n^2}=y^n\cdot \frac{1}{n^2}$$
As I said previously, you ignore the part with the variable raised to n, and just let the rest be $a_n$.
Another example to make it clear:
$$\sum \frac{x^n}{n3^n}=\sum x^n \cdot \frac{1}{n3^n}$$
so you let $a_n=\frac{1}{n3^n}$.

anonymous · Answer

ok thank you very much i think this gonna helps me

anonymous · Answer

sure i get it @blockcolder, believe me you will be very good professour insAllah ;)

blockcolder · Answer

Thanks for the complement. :D