Use $\sum_{n=0}^{\infty} \frac {2n +1} {2^n} = 6 $ to show $\sum_{n=0}^{\infty} \frac {2n +1} {2^n} i^n = \frac 4 {25} + i \frac {22} {25}$. I've shown that the series $$\sum_{n=0}^{\infty} \frac {2n +1} {2^n}$$ converges to $6$ using elementary series operations.
However how can I use this to show that $$\sum_{n=0}^{\infty} \frac {2n +1} {2^n} i^n = \frac 4 {25} + i \frac {22} {25}$$
and $$\sum_{p=0}^{\infty} (-1)^p\frac {4p +1} {4^p}$$ is convergent and has a sum which should be found ?
I don't see the connection, except that the second series is a product of a series I know is convervent (first series) and a series that is divergent.
 A: Hints:
$$i^n=\begin{cases}\;\;\,i&,\;\;n=1\pmod4\\-1&,\;\;n=2\pmod4\\-i&,\;\;n=3\pmod4\\\;\;\,1&,\;\;n=0\pmod4\end{cases}$$
Added on request: Using the above and the sum of a geometric series with ratio $\;r\;,\;\;|r|<1\;$
$$\sum_{k=0}^\infty\frac{2n+1}{2^n}i^n=2\sum_{n=0}^\infty \frac {i^nn}{2^n}+\sum_{n=0}^\infty\left(\frac i2\right)^n=$$
$$=2\sum_{n=0}^\infty\frac{4n+1}{2^{4n+1}}-2\sum_{n=0}^\infty\frac{4n+2}{2^{4n+2}}-2i\sum_{n=0}^\infty\frac{4n+3}{2^{4n+3}}+2\sum_{n=0}^\infty\frac{4n}{2^{4n}}+\frac1{1-\frac i2}=$$
Now use that (yes, again geometric, but this time power, series):
$$\frac1{1-x}=\sum_{n=0}^\infty x^n\implies\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}\;,\;\;|x|<1\;\;\;(**)$$
so for example
$$\sum_{n=0}^\infty\frac{4n+1}{2^n(=2\cdot2^{n-1})}=2\sum_{n=1}^\infty n\left(\frac12\right)^{n-1}+\frac1{1-\frac12}=2\frac1{\left(1-\frac12\right)^2}+2=10$$
$\color{red}{\text{But now I'm realizing there's a much easier way to do this...!}}$  
Using (**) above:
$$\sum_{k=0}^\infty\frac{2n+1}{2^n}i^n=i\sum_{n=0}^\infty n\left(\frac i2\right)^{n-1}+\sum_{n=0}^\infty\left(\frac i2\right)^n=i\frac1{\left(1-\frac i2\right)^2}+\frac1{1-\frac i2}=$$
$$=\frac{4i}{3-4i}+\frac2{2-i}=\frac{-16+12i}{25}+\frac{20+10i}{25}=\frac4{25}+\frac{22}{25}i$$
A: Cauchy-Hadamard formula shows the radius of convergence of the series $\sum\limits_{n=0}^n \frac{(2n+1)z^n}{2^n} $ is $2$. Techniques gives that series $=\frac{4+2z}{(z-2)^2}$ inside the disk(try it yourself! or at least verify it), and therefore if we take z=i the answer manifests. Sepereating the real part and the imaginary part gives the second answer.
A: Try to use the following:
$\sum_{n=0}^{\infty}{\frac{2n+1}{2^n}i^n} = \sum_{n=0}^{\infty}{(-1)^n\frac{4n+1}{4^n}} + \sum_{n=0}^{\infty}{(-i)^n\frac{2(2n+1)+1}{2^{2n+1}}}$
