OpenStudy (anonymous):

Find the interval of convergence of sqrt(1+x). Please help..Award At End.

OpenStudy (anonymous):

Do you have the power series for it?

OpenStudy (anonymous):

Yes I do @SithsAndGiggles my computer for some reason wont paste my work so I took a pic of it.

OpenStudy (anonymous):

I used Taylors Method to come to that answer.

OpenStudy (anonymous):

Alright, so let's use the ratio test: $\lim_{n\to\infty}\left|\frac{\dfrac{(-1)^{n+1}(2(n+1))!}{4^{n+1}((n+1)!)^2(1-2(n+1))}x^{n+1}}{\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)}x^n}\right|$ After some simplification... $\frac{|x|}{4}\lim_{n\to\infty}\frac{(2n+2)(2n+1)(2n-1)}{(n+1)^2(2n+1)}$

OpenStudy (anonymous):

I know I skipped quite a few steps, so some comments. Absolute value eliminates the powers of (-1). Then $$(2(n+1))!=(2n+2)!=(2n+2)(2n+1)(2n)!$$, and the factors of $$(2n)!$$ cancel out. $$\left|\dfrac{x^{n+1}}{x^n}\right|=|x|$$ and similarly, $$\left|\dfrac{\dfrac{1}{4^{n+1}}}{\dfrac{1}{4^n}}\right|=\dfrac{1}{4}$$. $$((n+1)!)^2=((n+1)n!)^2=(n+1)^2(n!)^2$$, and then you eliminate the $$(n!)^2$$ factors. $$|1-2(n+1)|=|1-2n-2|=|-1-2n|=2n+1$$. $$|1-2n|=|-1||2n-1|=2n-1$$, since $$2n-1>0$$ for $$n\to\infty$$.

OpenStudy (anonymous):

So the answer is 2n - 1 because 2n - 1 > 0 for n > inf

OpenStudy (anonymous):

The ratio test is where I made my mistake countless times...THANK YOU SO MUCH!

OpenStudy (anonymous):

After the simplification, you did more simplification didn't you? Because I noticed "left out some comments"

OpenStudy (anonymous):

Wait, what? "The answer is 2n-1 because..." No, the answer is whatever the interval of convergence is. I only said what to use (the ratio test) and skipped the algebra involved (which I tried to outline in my last comment).

OpenStudy (anonymous):

I'm sorry I miss understood you. Okay let me see...

OpenStudy (anonymous):

Let me just redo the simplifying step by step... \begin{align*} \lim_{n\to\infty}\left|\frac{\dfrac{(-1)^{n+1}(2(n+1))!}{4^{n+1}((n+1)!)^2(1-2(n+1))}x^{n+1}}{\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)}x^n}\right|&=\lim_{n\to\infty}\frac{\dfrac{(2(n+1))!}{4^{n+1}((n+1)!)^2|1-2(n+1)|}|x|^{n+1}}{\dfrac{(2n)!}{4^n(n!)^2|1-2n|}|x|^n}\\\\\\ &=\lim_{n\to\infty}\frac{\dfrac{(2(n+1))!}{4^{n+1}((n+1)!)^2(2n+1)}|x|^{n+1}}{\dfrac{(2n)!}{4^n(n!)^2(2n-1)}|x|^n}\\\\\\ &=\lim_{n\to\infty}\frac{\dfrac{(2n+2)(2n+1)(2n)!}{4^{n+1}(n+1)^2(n!)^2(2n+1)}|x|^{n+1}}{\dfrac{(2n)!}{4^n(n!)^2(2n-1)}|x|^n}\\\\\\ &=\lim_{n\to\infty}\frac{\dfrac{(2n+2)(2n+1)}{4(n+1)^2(2n+1)}|x|}{\dfrac{1}{(2n-1)}}\\\\\\ &=\frac{|x|}{4}\lim_{n\to\infty}\frac{\dfrac{(2n+2)(2n+1)}{(n+1)^2(2n+1)}}{\dfrac{1}{(2n-1)}}\\\\\\ &=\frac{|x|}{4}\lim_{n\to\infty}\dfrac{(2n+2)(2n+1)(2n-1)}{(n+1)^2(2n+1)}\\\\\\ \end{align*}

OpenStudy (anonymous):

Sorry...So the integral of convergence is [0,1]

OpenStudy (anonymous):

Let's see... For the series to converge, you must have $\frac{|x|}{4}\lim_{n\to\infty}\frac{(2n+2)(2n+1)(2n-1)}{(n+1)^2(2n+1)}<1$ The limit is $$\dfrac{2\cdot2\cdot2}{1\cdot1\cdot2}=4$$, which gives you $\frac{4|x|}{4}=|x|<1$ Judging by the factor of $$x^n$$ in the series, it looks like it's a series about $$x=0$$. The radius of convergence is 1, which makes the interval of convergence $$(-1,1)$$.

OpenStudy (anonymous):

Ohh okay. I did my math wrong. I was having n=0.

OpenStudy (anonymous):

That does make more sense. Thank you again. That helps and explains a lot.

OpenStudy (anonymous):

yw