<s>[INST] How do you calculate the expected value of geometric distribution without diffrentiation? Is there any way I can calculate the expected value of geometric distribution without diffrentiation? All other ways I saw here have diffrentiation in them.
Thanks in advance!
 [/INST] I know at least two ways off hand and there are probably others.
First I'll show you a concrete way to do it. After that I'll show you how to express the same thing exactly. (Together these make up only one of those "two ways". The other one now appears in the answer posted by "Henry".)
$$
\begin{array}{cccccccccccccccccccccccc}
& 0 & & 1 & & 2 & & 3 & & 4 & & 5 & & 6 \\
\hline
& & & p^1 & + & 2p^2 & + & 3p^3 & + & 4p^4 & + & 5p^5 & + & 6p^6 & + & \cdots & {} \\[12pt] 
= & & & p^1 & + & p^2 & + & p^3 & + & p^4 & + & p^5 & + & p^6 & + & \cdots \\
& & & & + & p^2 & + & p^3 & + & p^4 & + & p^5 & + & p^6 & + & \cdots \\
& & & & & & + & p^3 & + & p^4 & + & p^5 & + & p^6 & + & \cdots \\
& & & & & & & & + & p^4 & + & p^5 & + & p^6 & + & \cdots \\
& & & & & & & & & & + & p^5 & + & p^6 & + & \cdots \\
& & & & & & & & & & & & + & p^6 & + & \cdots \\
& & & & & & & & & & & & & & + & \cdots \\
& & & & & & & & & & & &  & & \vdots
\end{array}
$$
First sum each (horizontal) row. Each is a geometric series. Then sum the remaining series, which is also geometric.
Here is the same method expressed abstractly in the language of algebra:
\begin{align}
\sum_{x=0}^\infty x (1-p) p^x & = \sum_{x=1}^\infty x (1-p) p^x = \sum_{x=1}^\infty \sum_{j=1}^x (1-p)p^x \\[10pt]
& = \sum_{ x,j\, : \, 1 \,\le\, j \, \le \, x} (1-p) p^x = \sum_{j=1}^\infty \sum_{x=j}^\infty (1-p)p^x
\end{align}
Now you're summing a geometric series as $x$ goes from $j-1$ to $\infty,$ and then the outer sum, as $j$ goes from $1$ to $\infty,$ also turns out to be geometric.
(In the very first step above I put $\displaystyle\sum_{x=0}^\infty = \sum_{x=1}^\infty.$ That is justified by the fact that when $x=0,$ the actual term being added is $0$ so it can be dropped.)
</s>