Finding the sum $\sum_{n=1}^\infty \frac{(-1)^nx^{2n+1}}{4^{2n}(2n+1)}$

Question

Could someone please give me a hand with the following series?

$$\sum_{n=1}^\infty \frac{(-1)^nx^{2n+1}}{4^{2n}(2n+1)}$$

Answer 1

Let $f(x) =\sum_{n=1}^{∞} \frac{(-1)^nx^{2n+1}}{4^{2n}(2n+1)}$, so $f'(x) =\sum_{n=1}^{∞} \frac{(-1)^nx^{2n}}{4^{2n}}$ so the derivative converges iff $|x|< 4$. In this case $f'(x) =\frac{-x^2}{x^2+16}$. Integrating it you will get that $f(x) =4\tan^{-1}(\frac{x}{4}) -x$. You just need to analyse the limit cases for the serie with Leibniz's criterion

Answer 2

Given that : $$\sum _{n=1}^{\infty \:}\frac{\left(-1\right)^nx^{2n+1}}{4^{2n}\left(2n+1\right)}$$ $$$$ Use the Ratio Test to compute the convergence interval : $$\mathrm{Series\:Ratio\:Test:}$$ $$\mathrm{If\:there\:exists\:an\:}N\mathrm{\:so\:that\:for\:all\:}n\ge N,\:$$ $$a_n\ne 0\mathrm{\:and\:}\lim _{n\to \infty }|\frac{a_{n+1}}{a_n}|=L:$$ $$\mathrm{If\:}L<1\mathrm{,\:then\:}\sum a_n\mathrm{\:converges}$$ $$\mathrm{If\:}L>1\mathrm{,\:then\:}\sum a_n\mathrm{\:diverges}$$ $$\mathrm{If\:}L=1\mathrm{,\:then\:the\:test\:is\:inconclusive}$$ $$$$ Therefore : $$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{\frac{\left(-1\right)^{\left(n+1\right)}x^{2\left(n+1\right)+1}}{4^{2\left(n+1\right)}\left(2\left(n+1\right)+1\right)}}{\frac{\left(-1\right)^nx^{2n+1}}{4^{2n}\left(2n+1\right)}}\right|$$ $$$$ Compute : $$L=\lim _{n\to \infty \:}\left(\left|\frac{\frac{\left(-1\right)^{\left(n+1\right)}x^{2\left(n+1\right)+1}}{4^{2\left(n+1\right)}\left(2\left(n+1\right)+1\right)}}{\frac{\left(-1\right)^nx^{2n+1}}{4^{2n}\left(2n+1\right)}}\right|\right)$$ $$$$ Simplify : $$\frac{\frac{\left(-1\right)^{\left(n+1\right)}x^{2\left(n+1\right)+1}}{4^{2\left(n+1\right)}\left(2\left(n+1\right)+1\right)}}{\frac{\left(-1\right)^nx^{2n+1}}{4^{2n}\left(2n+1\right)}}:\quad -\frac{x^2\left(2n+1\right)}{16\left(2n+3\right)}$$ $$$$ Thus : $$L=\lim _{n\to \infty \:}\left(\left|-\frac{x^2\left(2n+1\right)}{16\left(2n+3\right)}\right|\right)$$ $$L=\left|-\frac{x^2}{16}\right|\cdot \lim _{n\to \infty \:}\left(\left|\frac{2n+1}{2n+3}\right|\right)$$ $$$$ We know that : $$\lim _{n\to \infty \:}\left(\left|\frac{2n+1}{2n+3}\right|\right)=1$$ $$$$ So : $$L=\left|-\frac{x^2}{16}\right|\cdot \:1$$ $$$$ Simplify : $$L=\frac{\left|x\right|^2}{16}$$ $$\mathrm{The\:power\:series\:converges\:for\:}L<1$$ $$\frac{\left|x\right|^2}{16}<1$$ $$$$ Find the radius of convergence : $$\frac{\left|x\right|^2}{16}<1:\quad \left|x\right|<4$$ $$\mathrm{Radius\:of\:convergence\:is}\:4$$ $$$$ Find the interval of convergence : $$\frac{\left|x\right|^2}{16}<1\quad :\quad -4<x<4$$ $$\mathrm{Check\:the\:interval\:end\:points}:\quad x=-4:\mathrm{converges},\:x=4:\mathrm{converges}$$ $$\mathrm{Interval\:of\:convergence\:is}\:-4\le \:x\le \:4$$ $$$$ Final answer is : $$\mathrm{Radius\:of\:convergence\:is}\:4,\:\:\mathrm{Interval\:of\:convergence\:is}\:-4\le \:x\le \:4$$