Convergence of series $x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots$

Question

Convergence of series $$S=x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots$$

When $x=\frac{1}{2}$

I used ratio test as:

$$a_{n}=\frac{\left(\frac{1}{2}\right)^{2n-1}}{2n-1}$$

Then $$a_{n+1}=\frac{\left(\frac{1}{2}\right)^{2n+1}}{2n+1}$$

Then we get:

$$\frac{a_{n+1}}{a_n}=\frac{2n-1}{4(2n+1)}$$

Hence we get:

$$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=0.25 \lt 1$$

Hence $\sum a_n$ Converges.

But how to find its value, i guess $\arctan x$ is not useful here

Answer 1

Take derivative of $S$ to get $$S'=1+x^2+x^4+...=\frac {1}{1-x^2}$$

Integrating and you get $$S=(1/2)\ln\left|\frac {1+x}{1-x}\right|$$

At $x=1/2$ the result is $\frac {\ln 3}{2}$

Answer 2

$$\log(1+x)=x-x^2/2-x^3/3+x^/4-....ad-inf, \log(1-x)=-x-x^2/2-x^3/3-.ad-inf~, if~ |x|<1.$$ Th given series ie quesion is nothing but $\frac{1}{2}\log[(1+x)/(1-x)]$ which is valid (convergent) for $|x|<1$.

Answer 3

If you take the derivative term by term, you get $1+x^2+x^4+\cdots=1/(1-x^2)$ (this is just a geometric series, convergent for $|x|<1$). Then you just integrate back the function $f(x)=1/(1-x^2)$ (using the trivial values $0$ at $0$) and you get the solution given by Dr Zafar Ahmed DSc.

Answer 4

You could aswell use the sandwich-lemma:

$$0 < x + \dfrac{x^3}{3} + \dfrac{x^5}{5} + ... < x+x^3+x^5+...<x+x^2+x^3+x^4+...=\sum_n^\infty 1\cdot x^n = \dfrac{1}{1-x}=2$$

Furthermore note that the sum is strictly increasing, hence the sum is convergent.