Calculate limit on series with nested sum I want to calculate the limit of following series:
$$\sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{1}{3^k} \cdot \frac{1}{2^{n-k}}$$
As far I could simply the series to:
$$\sum_{n=0}^{\infty} (\sum_{k=0}^{n} (\frac{1}{3})^k) \cdot (\sum_{k=0}^{n} 2^{n-k})$$
which would then allow me to use the geometric series:
$$\sum_{n=0}^{\infty} (\frac{1-(\frac{1}{3})^{n+1}}{\frac{2}{3}}) \cdot (\sum_{k=0}^{n}2^{n-k})$$
which can even be simplified further to:
$$\sum_{n=0}^{\infty} ((1-(\frac{1}{3})^n\cdot\frac{1}{3})(\frac{3}{2})) \cdot (\sum_{k=0}^{n}2^{n-k})$$
$$\sum_{n=0}^{\infty} (\frac{3}{2}-(\frac{1}{3})^n\cdot\frac{1}{2}) \cdot (\sum_{k=0}^{n}2^{n-k})$$
$$\sum_{n=0}^{\infty} (\frac{1}{2}(3-(\frac{1}{3})^n) \cdot (\sum_{k=0}^{n}2^{n-k})$$
This is as far as I know what to do. The solution by the way is:
$$\sum_{k=0}^{\infty} (\frac{1}{3})^k \sum_{k=0}^{\infty} (\frac{1}{2})^k$$ which could be simplified again with the geometric series.
However I have now idea what to do with $\sum_{k=0}^n2^{n-k}$ though we could write it as $\sum_{k=0}^n 2^n \sum_{k=0}^n 2^{-k}$ this would not make much sense as the former sum would converge against infinity.

$$\sum^{\infty}_{n=0}\sum^{n}_{k=0}\frac{1}{3^k}\cdot\frac{1}{2^{n-k}}=\sum^{\infty}_{n=0}\sum^{n}_{k=0}\frac{1}{3^k}\cdot\frac{2^k}{2^{n}}=\sum^{\infty}_{n=0}\frac{1}{2^{n}}\sum^{n}_{k=0}\Big(\frac{2}{3}\Big)^k$$
Using the geometric series result for a convergent geometric series yields
$$\sum^{\infty}_{n=0}\frac{1}{2^{n}}\sum^{n}_{k=0}\Big(\frac{2}{3}\Big)^k=\sum^{\infty}_{n=0}\frac{1}{2^{n}}\cdot\frac{1-\Big(\frac{2}{3}\Big)^{n+1}}{1-\frac{2}{3}}=3\cdot\sum^{\infty}_{n=0}\frac{1-\Big(\frac{2}{3}\Big)^{n+1}}{2^{n}}=3\cdot\sum^{\infty}_{n=0}\Big(\frac{1}{2^n}-\frac{2}{3}\cdot\frac{1}{3^n}\Big)=3\cdot\sum^{\infty}_{n=0}\frac{1}{2^n}-2\cdot\sum^{\infty}_{n=0}\frac{1}{3^n}=3\cdot\frac{1}{1-1/2}-2\cdot\frac{1}{1-1/3}=6-3=3$$
