calculating an infinte sum given that : $\alpha_{i}$ = $\frac1{2^{i+1}}$  and  $\beta_{i}$ = (1+$\frac{{(i-1)i}}{2} +(2i))$ 2
How can we calculate this infinite  sum :
$$\sum_{k=1}^\infty{ \alpha_{i}\beta_{i}}$$
I calculate 1200 terms of this sum using python , and it is seems that it converges to 8 .
How can we proove that mathematically ?
This was my try :
$$\sum_{k=1}^\infty{ \alpha_{i}\beta_{i}}$$ $$=\sum_{k=1}^\infty{ \left(\frac1{2^{i}} + \frac{{(i-1)}i}{2^{i+1} } + \frac{{2}i}{2^{i} }\right)}$$
The first term is easy ( geometric serie ), but what the second and the third term ?
 A: You seem to indicate from your question that you understand how to calculate the first term. As you mentioned, it's a geometric series, the formula for which is well known. The last term is a little more complicated. It's an arithmetico-geometric sequence. The formulas for these are known, and can be found in the linked article:
$$\Sigma i(r)^i = \frac{r}{(1 - r)^2}$$
So for $r = \frac12$ we have:
$$\Sigma i(\frac12)^i = \frac{\frac12}{(\frac12)^2} = \frac{\frac12}{(\frac14)} = 2$$
The middle term is yet again slightly more complicated. Let's work out a general formula first that'll help us. Suppose we want to calculate the infinite sum $S$ given below:
$$S = \Sigma i^2r^i$$
Assume that $r$ is such that the sum converges. Then we have can algebraically manipulate it to get an identity as follows:
$$S(1 - r) = S - rS = \Sigma i^2r^i - \Sigma i^2r^{i+1} = \\
\Sigma i^2r^i - \Sigma (i - 1)^2r^i = \\
\Sigma i^2r^i - (i - 1)^2r^i = \Sigma (2i - 1)r^i = \\
2\Sigma ir^i - \Sigma r^i$$
Now we've reduced to terms we already know: a geometric series plus an arithmetico-geometric series. Applying the formulas we've already seen for those:
$$ = 2\left(\frac{r}{(1 - r)^2}\right) - \frac{r}{1 - r}$$
Remember this is the term for $S(1 - r)$ so to get $S$ we divide by $1 - r$ to get:
$$S = 2\left(\frac{r}{(1 - r)^3}\right) - \frac{r}{(1 - r)^2}$$
In particular, when $r = \frac12$ then $S$ is:
$$S = 2\left(\frac{\frac12}{(\frac12)^3}\right) - \frac{\frac12}{( \frac12)^2} = \\
2\left(\frac{\frac12}{(\frac18)}\right) - \frac{\frac12}{(\frac14)} = \\
2(\frac82) - \frac42 = 8 - 2 = 6$$
You now have mathematical formulae for all the terms in your expression. Let's plug them into your expression and just use some algebra/rearrangement of sums to work out the result. First, though, I think your formula is slightly incorrect. You have:
$$\sum{ \left(\frac1{2^{i}} + \frac{{(i-1)}i}{2^{i+1} } + \frac{{2}i}{2^{i} }\right)}$$
But I believe you should start with:
$$\sum{ \left(\frac2{2^{i}} + \frac{{(i-1)}i}{2^{i+1} } + \frac{{2}i}{2^{i} }\right)}$$
Notice the $2$ multiplier on the first term, which I think you forgot. From here, you can derive your convergence to $8$:
$$\sum{ \left(\frac2{2^{i}} + \frac{{(i-1)}i}{2^{i+1} } + \frac{{2}i}{2^{i} }\right)} = \\
2\sum\frac{1}{2^{i}} + \sum\frac{{(i-1)}i}{2^{i+1} } + \sum\frac{{2}i}{2^{i} } = \\ 2(1) + \sum\frac{i^2}{2^{i+1} } - \sum\frac{i}{2^{i+1} } + 2\sum\frac{i}{2^{i} }
= \\ 2 + \frac12\sum\frac{i^2}{2^i } - \frac12\sum\frac{i}{2^i} + 2\sum\frac{i}{2^{i} }
= \\ 2 + \frac12\sum\frac{i^2}{2^i } + \frac32\sum\frac{i}{2^{i} } \\
= \\ 2 + \frac12\sum i^2(\frac{1}{2})^i + \frac32\sum i(\frac{1}{2})^i \\
= \\ 2 + \frac12(6) + \frac32(2) = \\
2 + 3 + 3 = 8$$
