Expectation of Geometric distribution problem 
Find the expectation of a Geometric distribution using $\mathbb{E}(X)= \sum_{k=1}^\infty P(X \ge k)$. 

Okay I know how to find the expectation using the definition of the geometric distribution $$P(X=k)= p \cdot(1-p)^{k-1}$$ and I figured that $P(X \ge k)=(1-p)^{k-1}$ but I don't know how to show it. 
I know the expectation is $\frac{1}{p}$ but I just get $\mathbb E(X)= \frac{1}{p^2}$ using the method specified in the question.
 A: For $|r|&lt1$, the sum of the geometric series $\sum\limits_{k=1}^\infty r^k$ is ${ r\over 1-r}$.
So, write
$$\sum\limits_{k=1}^\infty P[X\ge k]= \sum\limits_{k=1}^\infty (1-p)^{k-1}
= {1\over 1-p}\sum\limits_{k=1}^\infty (1-p)^{k },$$
and apply the formula with $r=1-p$.
A: The answer that is here does not address one aspect of the question:

I figured that $P(X \ge k)=(1-p)^{k-1}$ but I don't know how to show it.[...]

Here is an hint: 
$$P(X \ge k)=\sum_{i=k}^\infty P(X=i)$$
Now, to evaluate the above sum, you need the sum of the geometric series: 

For $|r|&lt1$, $$\sum_{i=k}^\infty r^i=\frac{ r^k }{1-r}$$

The rest of the details are there in David's answer...but in case you need to know more about one or more of this, you may want to ping me here...
A: A simpler way would be to plug in $q=1-p$ and solve it that way using formula for geometric sequences:
\begin{align*}
E(X) &= \sum\limits_{k=1}^\infty kpq^{k-1}\\
&= \frac{p}{q} \sum\limits_{k=1}^\infty kq^{k}\\
&= \frac{p}{q} \frac{q}{(1-q)^2}\\
&= \frac{p}{q} \frac{q}{p^2}\\
&= \frac{p}{q} \frac{q}{p^2}\\
&= \frac{1}{p}.
\end{align*}
