Determine the convergence of interval for the function $\tan^{-1}x$ . I know this question is very classical and standard and I am not very sure of the part for the end points .
For $x\in(-1,1)~$ , one has $$\tan^{-1}x=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}x^{2n+1}~.$$
Determine the end points $x=\pm 1$ ,
$(i)~$ for $x=1$ , the summation turns into $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}$ , then this is an alternating series since each tern $\displaystyle\frac{1}{2n+1}$ decreases and $\displaystyle\lim_{n\rightarrow\infty}\frac{(-1)^{n}}{2n+1}=0$ .
$(ii)$ for $x=-1$ , it becomes $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}(-1)^{2n+1}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}(-1)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{2n+1}$ , thus , this series converges by alternating series test since for the similar reasons that each term $\displaystyle\frac{1}{2n+1}$ is decreasing and $\displaystyle\lim_{n\rightarrow\infty}\frac{(-1)^{n+1}}{2n+1}=0$
Can someone check my proof for validity if you have the time . Any comment or advice I will be grateful .
Looks good.
I know you see the denominator goes to infinity, but it's more clear and useful to do it like so when dealing with more tedious coefficients
$$\left|\lim_{n\to\infty}\frac{(-1)^n}{2n+1}\right|=\lim_{n\to\infty}\frac{1}{2n+1}=\lim_{n\to\infty}\frac{n\frac{1}{n}}{n\left(2+\frac{1}{n}\right)}=\lim_{n\to\infty}\frac{\frac{1}{n}}{2+\frac{1}{n}}=0$$
It would also be better to get in the practice of proving that the sequence decreases for the same reasons like this
$\displaystyle a_n=\frac{1}{2n+1}\sim f(x)=\frac{1}{2x+1}$
$$\frac{df}{dx}=\frac{-2}{(2x+1)^2}<0\space\forall\space x\in\mathbb{R^+}$$
$$\therefore a_n\space\text{decreases for all n}$$
This is typical but certainly not always the case