I need know how to do, please

Question

loser66 · Answer

attachment

loser66 · Answer

I need know whether they are converge or not and their radii., please

hartnn · Answer

@KingGeorge

kinggeorge · Answer

My first thought, is to just let $n\to\infty$ and see what happens. In this case, you have$$\frac{a_n(n^2+2n-1)}{2n^2+6n+4}$$As n goes to infinity, you then get$$\frac{a_n}{2}.$$While this doesn't prove convergence, it seems to imply that it probably does converge since as n increases, your $a_n$ should always be decreasing.

kinggeorge · Answer

Sorry, missed a negative. Should be$$-\frac{a_n}{2}$$But if you take the absolute value, you get an always decreasing sequence.

loser66 · Answer

from this, I can have 2 solutions
$$y_1 = 1+\frac{1}{4}x^2 -\frac{7}{96}x^4+.....$$
and $$y_2 = x -\frac{1}{6}x^3+\frac{7}{120 }x^5 -...$$
by letting a_0 =0 and a_1 =1 and letting a_0 =1 and a_1 =0
so for those series, how to consider radii and convergences?

kinggeorge · Answer

Since we have an alternating series, you might consider using the alternating series test. It's pretty easy to see that your terms are always decreasing, and as $n\to\infty$ this approaches just dividing by 2 every time, so it clearly goes to 0 as $n\to\infty$. So it must converge. Since $a_n$ is completely arbitrary in this, the radius of convergence should be all real numbers.

loser66 · Answer

should we have any formula to consider it?

kinggeorge · Answer

There's probably some formula, but I'm not sure what it is. Also, I'm reconsidering my method of finding the radius of convergence. Give me a minute to revise it.

loser66 · Answer

thanks for being here.

kinggeorge · Answer

Upon further review, the radius of convergence is not all real numbers. To see what it actually is, we apply the ratio test. For our purposes, we will approximate our series by$$a_n=\frac{a_0}{2^n}$$Applying the ratio test,$$\lim_{n\to\infty}\left|\frac{a_0/2^{n+1} }{a_0/2^n}\frac{x^{n+1}}{x^n}\right|=\lim_{n\to\infty}\left|\frac{x}{2}\right|=\left|\frac{x}{2}\right|$$For this to be less than 1, we need $|x|<2$. So that should be the radius of convergence.

kinggeorge · Answer

We can approximate our series by that since it approaches that series $n\to\infty$, and small values of $n$ really don't have that much effect.

loser66 · Answer

is there something wrong? $\huge a_n \neq \frac{a_0}{2^n}$

kinggeorge · Answer

I did a little bit of a simplification so we didn't have messier exponents. Basically I just defined a new series by $a_{n+1}=a_n/2$, which closely approximates the original series. 

But you're right, something did go wrong. But it's with the exponents of $x$, not the series. Our new ratio test should be $$\lim_{n\to\infty}\left|\frac{a_0/2^{n+1} }{a_0/2^n}\frac{x^{2n+2}}{x^{2n}}\right|=\lim_{n\to\infty}\left|\frac{x^2}{2}\right|=\frac{x^2}{2}$$So if this is less than 1, then we see that $x^2<2$ so $|x|<\sqrt2$.

loser66 · Answer

I got this part, this for series of even indices, right?

kinggeorge · Answer

Correct, and if you have the odd indices, you start slightly differently, but you should still get the same thing (in this case).

loser66 · Answer

really? ok, let me try, Thanks aaa bomb.

kinggeorge · Answer

no problem.