MHB Proof of tan^(-1z) maclauren series

[Stumped1]

prove that
$$tan^{-1}z=z-z^3/3 + z^5/5 -z^7/7 + ...$$ for $$|z|<1$$

I know of a proof for this that takes the derivative, does long division, then integrates.

I would like a proof of this using the known Maclaurin series for e^z, cosz, or sinz.Is there a way to do this using these?

Thanks for any help!

 

[Prove It]

prove that
$$tan^{-1}z=z-z^3/3 + z^5/5 -z^7/7 + ...$$ for $$|z|<1$$

I know of a proof for this that takes the derivative, does long division, then integrates.

I would like a proof of this using the known Maclaurin series for e^z, cosz, or sinz.Is there a way to do this using these?

Thanks for any help!

The easiest way is remember that $\displaystyle \begin{align*} \frac{d}{dx} \left[ \arctan{(x)} \right] = \frac{1}{1 + x^2} \end{align*}$.

Now notice $\displaystyle \begin{align*} \frac{1}{1 + x^2} = \frac{1}{1 - \left( -x^2 \right) } \end{align*}$, and if we recall that a geometric series has $\displaystyle \begin{align*} \sum_{n = 0}^{\infty} r^{n} = \frac{1}{1 - r} \end{align*}$ if $\displaystyle \begin{align*} |r| < 1 \end{align*}$, that means $\displaystyle \begin{align*} \frac{1}{1 - \left( - x ^2 \right) } = \sum_{n = 0}^{\infty} { \left( -x^2 \right) ^n } \end{align*}$ if $\displaystyle \begin{align*} \left| -x ^2 \right| < 1 \end{align*}$. Thus

$\displaystyle \begin{align*} \frac{1}{1 + x^2} &= \sum_{n = 0}^{\infty}{ \left( -1 \right) ^n x^{2n} } \textrm{ if } |x| < 1 \\ \int{ \frac{1}{1 + x^2} \, dx} &= \int{ \sum_{n = 0}^{\infty}{ \left( -1 \right) ^n x^{2n} }\,dx} \\ \arctan{(x)} + C &= \sum_{n = 0}^{\infty} \frac{ \left( -1 \right) ^n x^{2n + 1} }{2n + 1} \end{align*}$

and by substituting $\displaystyle \begin{align*} x = 0 \end{align*}$ it can be seen that $\displaystyle \begin{align*}C = 0 \end{align*}$.

Thus $\displaystyle \begin{align*} \arctan{(x)} = \sum_{n = 0}^{\infty} \frac{\left( -1 \right) ^n x^{2n + 1}}{2n + 1} \end{align*}$ if $\displaystyle \begin{align*} |x| < 1 \end{align*}$.