# Find the sum of power series $\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1}}{2n-1}$

Find the sum of power series $$\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1}}{2n-1}$$

How can we use derivation/integration method for this series?

$$\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1}}{2n-1}$$ is convergent for $-1<x<1$. Then, in case of $0<x<1$ and with $X=x^{1/2}$ Let: $$y(X)=\sum_{n=1}^{\infty}\frac{(-1)^nX^{2n-1}}{2n-1}$$ (wich is different from the above series. Do not confuse them. We will come back to the above series at the end). $$y'(X)=\sum_{n=1}^{\infty}(-1)^nX^{2n-2}=-\frac{1}{1+X^2} \qquad \text{geometric series}$$

$$Y(X)=-\int \frac{dx}{1+X^2}=-\tan^{-1}(X)+c$$ $X=0 \quad\to\quad Y(0)=\sum_{n=1}^{\infty}\frac{(-1)^n 0^{2n-1}}{2n-1}=0=-\tan^{-1}(0)+c \quad\to\quad c=0$ $$y(X)=\sum_{n=1}^{\infty}\frac{(-1)^nX^{2n-1}}{2n-1}=-\tan^{-1}(X)$$ With $X=x^{1/2}$ $$\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1/2}}{2n-1}=-\tan^{-1}(x^{1/2})$$ $$x^{-1/2}\sum_{n=1}^{\infty}\frac{(-1)^nx^{n}}{2n-1}=\tan^{-1}(x^{1/2})$$ $$\sum_{n=1}^{\infty}\frac{(-1)^nx^{n}}{2n-1}=-x^{1/2}\tan^{-1}(x^{1/2})$$ Do similar calculus in case of $-1<x<0$.

• If you don't want the OP to be confused do not use $\;X\;$ in your second-third line. Use $\;a,\,\alpha,\,\omega\;$ or something like that. There's no problem at all to differentiate/integrate wrt $\;a,\,\alpha,\,\omega\;$ orwhatever. – DonAntonio Feb 11 '17 at 11:12
• @JJacquelin, Why did we chose substitution $x^{1/2}$? – user300045 Feb 11 '17 at 12:54
• The denominator is $2n-1$. If we want to cancel it through differentiation, the power of $X$ must be $2n-1$. Since the actual power is $n-1$ we have to change $x^n$ into $X^{2n}$, so $x$ into $X^2$. – JJacquelin Feb 11 '17 at 14:05

For $x\ge0,$

$$\dfrac{(-1)^nx^{n-1}}{2n-1}=\dfrac ix\dfrac{(i\sqrt x)^{2n-1}}{2n-1}$$

Now $$\ln(1+y)-\ln(1-y)=\sum_{r=1}^\infty\dfrac{y^{2r-1}}{2r-1}$$

• Could you show more intuitive method? – user300045 Feb 11 '17 at 10:41
• @user_99, Not sure about intuitive term! Whenever the $n$ term of contains a linear expression of $n$ in the denominator , Log Series comes to my mind. Whenever the $n$ term of contains a factorial of $n$ in the denominator, e series. Whenever there is an Arithmetic Series in the numerator=> Binomial Series – lab bhattacharjee Feb 11 '17 at 10:51
