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

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:

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

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}

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:

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

• No need to involve imagine numbers when you can write the geometric series as $\sum_{n=1}^\infty(-x^2)^n$ no? – Simply Beautiful Art May 9 at 12:59
• Yep, it was late when I wrote it. Edited. – zwim May 9 at 19:00