Question: How we can show if the sum of $$\lim\_{n\rightarrow \infty }\sum\_{k=0}^{n} \frac{1}{{n \choose k}}$$ converges and then find the result of the sum if it converges?

Thanks for any help.

Answer: 
By the linearity of expectation, we have:
$$
E(Y) = E(-X + 1) = -E(X) + 1
$$

Alternatively, you could use the definition directly. Let $Y=g(X)=-X+1$. Then:
$$ \begin{align\*}
E(Y)&=E(g(X))\\
&=\int\_{-\infty}^\infty g(x)f(x)dx\\
&=\int\_{-\infty}^\infty(-x+1)f(x)dx\\
&=-\int\_{-\infty}^\infty xf(x)dx + \int\_{-\infty}^\infty f(x)dx\\
&=-E(X) + 1
\end{align\*}$$
You can calculate the expected value of some function $g$ of $X$ using the following formula:

$$E[g(X)]=\int\_{-\infty}^{\infty}g(x)f(x)\:dx$$

An explanation for this can be found on the [wikipedia page](http://en.wikipedia.org/wiki/Expected_value#General_definition).