Find an explicit formula for a function that is exactly equivalent to the power series $\sum_{n=1}^\infty\frac{(-1)^n}{2n+1}x^{2n}$

Question

Find an explicit formula for a function that is exactly equivalent to the power series $$\sum_{n=1}^\infty\frac{(-1)^n}{2n+1}x^{2n}$$

Can somebody give me an idea on where to start with this?

Edit: Thanks to everybody's comments, I recognize that this power series is similar to the one for arctan:

$$\tan^{-1}x=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}x^{2n+1}$$

Please let me know if what I did was correct:

I decided to try and get the $x^{2n}$ in the first power series to $x^{2n+1}$ by multiplying the power series by $\frac{x}{x}$ to get:

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

I then factored out $\frac{1}{x}$ to get:

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

Considering that $\sum_{n=1}^\infty a_n=\bigg(\sum_{n=0}^\infty a_n\bigg)-a_0$ this would mean that:

$$\frac{1}{x}\sum_{n=1}^\infty\frac{(-1)^n}{2n+1}{x^{2n+1}}=\frac{1}{x}\bigg(\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}x^{2n+1}-x\bigg)$$

Which makes the power series equivalent to:

$$\frac{1}{x}(\tan^{-1}x-x)$$

Answer 1

Your work is correct. As an alternative approach that avoids recognizing the Maclaurin series for $\arctan$, it is easy to note that $\frac1{2n+1}=\int_0^1t^{2n}~\mathrm dt$, hence with geometric series and the derivative of $\arctan$,

\begin{align}\sum_{n=1}^\infty\frac{(-1)^n}{2n+1}x^{2n}&=\int_0^1\sum_{n=1}^\infty(-1)^nx^{2n}t^{2n}~\mathrm dt\\&=\int_0^1\sum_{n=1}^\infty(-x^2t^2)^n~\mathrm dt\\&=\int_0^1\frac{-x^2t^2}{1+x^2t^2}~\mathrm dt\\&=\int_0^1\frac1{1+x^2t^2}-1~\mathrm dt\\&=\frac1x\arctan(x)-1\end{align}

Answer 2

To find a closed form for a power series it is often interesting to find a differential equation verified by the function. Not also that it works both ways, since given an ODE we can try to find a power series verifying it and that gives relations between coefficients.

Anyway, here the first thing to notice is that the $2n+1$ denominator can be cancelled by derivation of $x^{2n+1}$.

And since we only get $x^{2n}$ on numerator we ought to multiply the whole thing by $x$.

Thus $\quad xf(x)=x\sum\limits_{n=1}^{\infty} \dfrac{(-1)^n x^{2n}}{2n+1}=\sum\limits_{n=1}^{\infty} \dfrac{(-1)^n x^{2n+1}}{2n+1}$

Now by derivating $\quad (xf(x))'=\sum\limits_{n=1}^{\infty} (-1)^n x^{2n}=\sum\limits_{n=1}^{\infty} (-x^2)^n\quad$ and we recognize a geometric sum.

Till now, we haven't paid much attention to convergence, but obviously the radius of convergence is $1$ so we will assume $|x|<1$ from now on.

The summation gives $(xf(x))'=\underbrace{(-x^2)}_{\text{sum starts at n=1}}\dfrac {1-0}{1-(-x^2)}=\dfrac {-x^2}{1+x^2}=\dfrac{1}{1+x^2}-1\quad$

[since $(-x^2)^N\to 0$]

Integrating this ODE leads finally to $xf(x)=\arctan(x)-x+C$

The power series gives initial condition $f(0)=0\implies C=0$ and we have found the closed form:

$$f(x)=\dfrac{\arctan(x)}x-1$$