Question about a infinite series Could you please explain how come:
$$\sum_{i=1}^\infty(0.5)^{i+1}(i+1)=2$$
How come the answer is not inf? And why 2?
Any help will be appreciated.
 A: We know for $|r|<1$, $$\sum_{1\le i<\infty}r^{i+1}=r^2\cdot\frac1{1-r}=\frac{r^2}{1-r}$$
$$\text{Now, }\frac{d\left(\sum_{1\le i<\infty}r^{i+1}\right)}{dr}=\sum_{1\le i<\infty}\frac{d r^{i+1}}{dr}=\sum_{1\le i<\infty}(i+1)r^i$$
$$\text{and  }\frac{d\left(\frac{r^2}{1-r}\right)}{dr}=\frac{2r-r^2}{(1-r)^2}$$
$$\implies \sum_{1\le i<\infty}(i+1)r^i=\frac{2r-r^2}{(1-r)^2}$$
$$\implies \sum_{1\le i<\infty}(i+1)r^{i+1}=\frac{r(2r-r^2)}{(1-r)^2}$$
Alternatively, using Arithmetico-geometric series,  we can show, $$\sum_{1\le i<\infty}(i+c)r^{i+1}=\frac{(c+1)r^2-c\cdot r^3}{(1-r)^2}$$ where $c$ is any finite number
Let $S=\sum_{1\le i<\infty}(i+c)r^{i+1}$
Multiply either sides by the common ratio $r,$  and subtract from the original
A: You wonder why the sum
$$\sum_{k=2}^{\infty} k \left ( \frac{1}{2}\right)^k$$
does not blow up, presumably because of the factor of $k$ in the sum which grows linearly to $\infty$.  The answer, very roughly, is because the factor of $(1/2)^k$ decreases much more quickly than $k$ increases.
Let's look at a few terms to see how this works.  The first few terms in the sum look like
$$2 \left( \frac{1}{2}\right)^2 + 3 \left( \frac{1}{2}\right)^3 + 4 \left( \frac{1}{2}\right)^4 + \ldots = \frac{2}{4} + \frac{3}{8} + \frac{4}{16}+ \ldots$$
The terms here are already getting smaller at $k=2$.  Now let's double the index:
$$\ldots+\frac{8}{2^8}+\ldots+\frac{16}{2^{16}}+\ldots+\frac{32}{2^{32}}+\ldots$$
You should see that, while the numerator doubles, i.e., exponent increases by $1$, the denominator is squared, i.e., exponent is doubled. 
I hope you see that the terms in the series rapidly vanish as $k \rightarrow \infty$.  By comparison with far more slowly varying sequences whose corresponding series converge (e.g., $1/k^2$), this series converges.
It converges to $3/2$ for reasons summarized by @lab above.
