Convergence of series of functions

Question

$$\sum_{n=0}^{\infty} \left(\dfrac{x^{2n+1}}{2n+1}-\dfrac{x^{n+1}}{2n+2}\right) x\in[0;1]$$ Could I find the convergence of this series of functions using derivatives?

Answer 1

For $x\in(0,1)$,

$$F(x)=\sum_{n=0}^{+\infty} \frac{x^{2n+1}}{2n+1}-G(\sqrt{x})$$ with $$G(x)=\sum_{n=0}^{+\infty}\frac{x^{2n+2}}{2n+2}$$

$$G'(x)=\frac{x}{1-x^2}$$

$$\implies G(x)=-\frac{1}{2}\ln(1-x^2)$$

$$F(x)=\frac{1}{2}\ln(\frac{1+x}{1-x})+\frac{1}{2}\ln(1-x).$$

$$=\frac{1}{2}\ln(1+x).$$

$F(0)=0$

Answer 2

For any $x\in(-1,1)$ we have: $$ \sum_{n\geq 0}\frac{x^{2n+1}}{2n+1} = \text{arctanh}(x) = \frac{1}{2}\log\frac{1+x}{1-x},\qquad \sum_{n\geq 0}\frac{x^{n+1}}{2n+2} = -\frac{1}{2}\log(1-x) $$ by well-known Taylor series. It follows that, for any $x\in(-1,1)$, $$ \sum_{n\geq 0}\left(\frac{x^{2n+1}}{2n+1}-\frac{x^{n+1}}{2n+2}\right) = \log\sqrt{1+x}. $$ Equality does not hold for $x=1$: $$ \sum_{n\geq 0}\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right) = \sum_{n\geq 0}\frac{1}{(2n+1)(2n+2)}=\int_{0}^{1}\frac{dx}{1+x}=\log(2). $$

Answer 3

As you know, the geometric series $\sum_{n=0}^{\infty} x^n$ converges for $x \in [0,1).$ Thus each of the series $\sum_{n=0}^{\infty}x^{2n+1}/(2n+1),$ $\sum_{n=0}^{\infty}x^{n+1}/(2n+2)$ converges for $x \in [0,1)$ by the comparison test. Hence so does your series for these $x.$

For $x = 1,$ your series is

$$\sum_{n=0}^{\infty}\left ( \frac{1}{2n+1} - \frac{1}{2n+2}\right ) = \sum_{n=0}^{\infty} \frac{1}{(2n+1)(2n+2)}.$$

The series on the right can be compared to $\sum_{n=1}^{\infty}1/n^2$ to show convergence.

Conclusion: Your series converges for all $x\in [0,1].$