Determine the value of a series

Question

The series is:

$\sum_{n=1}^\infty 2n (-\frac{4}{9})^n x^{2n-1}$

I think I have to simplify it to have a cosinus and something else. I tried a few times but I just cant get it to: something + or * $\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$

Answer 1

For the first, use How I can calculate $\sum_{n=0}^\infty{\frac{n}{3^n}}$?

Observe that $$\sum_{n=1}^\infty 2n \left(-\frac49\right)^n x^{2n-1}=-\frac{4x}9\cdot\frac2x\sum_{n=1}^\infty n \left(-\frac{4x}9\right)^{n-1}$$

Check for the convergence of $\displaystyle\sum_{n=1}^\infty \left(-\frac{4x}9\right)^n$

For the second $$\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}=\sum_{n=0}^\infty \frac{(ix)^{2n}}{(2n)!}$$

Now from this defintion $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\ \ \ \ (1)$$

$$\implies e^{-x}=\sum_{n=0}^\infty\frac{(-x)^n}{n!}\ \ \ \ (2)$$

Check $(1)+(2)$

Answer 2

Hint: it's the derivative of a geometric series.