Why does this equivalence of sums hold? I have to prove that a function is a pdf. In the master solution they state the below equivalence in the proof. What rules are applied to get from the left side to the right side?
$$C\sum_{k=2}^{\infty} \sum_{j=1}^{k-1}\left(\frac{1}{2}\right)^{k}=C \sum_{j=1}^{\infty} \sum_{k=j+1}^{\infty}\left(\frac{1}{2}\right)^{k}$$

Using geometric series, for the inner sum we have $\sum_{j=1}^{k-1}\left(\frac{1}{2}\right)^k=\frac{k-1}{2^k}$. Now inserting this result into the outer sum, we get $\sum _{k=2}^R \frac{k-1}{2^k}=1-\frac{R+1}{2^R}$.
Putting this together, we obtain for the left side (by using the limit $R$ instead of infinity):
$$\sum_{k=2}^{R}\sum_{j=1}^{k-1}\left(\frac{1}{2}\right)^k=1-\frac{R+1}{2^R}$$
So what is happening on the right side? For the inner sum we have $\sum _{k=j+1}^R \left(\frac{1}{2}\right)^k=\frac{1}{2^j}-\frac{1}{2^R}$. Let us insert this result into the outer sum and we obtain as well $\sum _{j=1}^R \left(\frac{1}{2^j}-\frac{1}{2^R}\right)=1-\frac{R+1}{2^R}$.
Therefore, when replacing on the right side in both sums infinity with $R$, you get as well:
$$\sum_{j=1}^{R}\sum_{k=j+1}^{R}\left(\frac{1}{2}\right)^k=1-\frac{R+1}{2^R}$$
Hence we can conclude $C\left(1-\frac{R+1}{2^R}\right)=C\left(1-\frac{R+1}{2^R}\right)$ which is true.
