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.

You have terms like $(\frac{1}{2^n} + \frac{1}{3}\cdot \frac{1}{2^{n-1}} + \cdots + \frac{1}{3^n})$ for each $n$. We can collect terms in $n$ instead of $k$ - we get a series $\frac{1}{3^k} \sum_{i = 0}^\infty 2^{-i}$ for each $k$.
E.g., pick a value of $k$. For $\frac{1}{3^k}$, there exists exactly one term also containing $2^{n-k}$ for any one given positive $n-k$. Collect all these terms to create the sum. We get
$$
\sum\sum(\cdots) = \sum_{k = 0}^\infty \frac{1}{3^k}\sum_{n=0}^\infty \frac{1}{2^n} \\
= 2\sum_{k = 1}\frac{1}{3^k} \\
= 2 \frac{1}{1 - \frac{1}{3}} \\
= 3
$$
