Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
 A: Such a sequence is called Arithemtico-Geomteric Progression.
$$S_n=\sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i } }  } $$
$$\frac{S_n}{2}=\sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i+1 } }  }=\sum _{ i=2 }^{ n+1 }{ \frac { i-1 }{ { 2 }^{ i } }  }$$
Subtracting
$$\frac{S_n}{2}=\sum _{ i=1 }^{ n }{ \frac { 1 }{ { 2 }^{ i } }  } -\frac { n }{ { 2 }^{ n+1 } }   $$
as $n\rightarrow \infty$
It's easily seen that $S_{\infty}=2$

How to evaluate that limit
$$\sum _{ i=1 }^{ n }{ x^{i-1}  } =\frac{1-x^n}{1-x}$$ If $|x|<1$ as $n\rightarrow \infty$
$$\sum _{ i=1 }^{ n }{ x^{i-1}  } =\frac{1-x^n}{1-x}=\frac{1}{1-x}$$
Second one is directly from Taylor series.

Although there exist simpler proof , I have a rigorous proof of the second part of the limit
$$0<\log _{ 2 }{ x } =\log _{ 2 }{ e } \int _{ 1 }^{ x }{ \frac { 1 }{ t }  } dt$$
When $x>1$ $\frac{1}{t}<\frac{1}{\sqrt{t}}$ is valid
$\log _{ 2 }{ e } \int _{ 1 }^{ x }{ \frac { 1 }{ t }  } dt<\log _{ 2 }{ e } \int _{ 1 }^{ x }{ \frac { 1 }{ \sqrt { t }  }  } dt=2\log _{ 2 }{ e } \left( \sqrt { x } -1 \right) <2\log _{ 2 }{ e }\cdot \sqrt { x } $
$0<\log _{ 2 }{ x } <2\log _{ 2 }{ e } \sqrt { x }$ $$ \Rightarrow 0<\frac { \log _{ 2 }{ x }  }{ x } <\frac { 2\log _{ 2 }{ e }  }{ \sqrt { x }  } \tag{1} $$
As $x\rightarrow \infty$ using $(1)$ and squeeze principle. We get
$$\lim_{x\rightarrow \infty}{\frac{\log_{2}{x}}{x}}=0\tag{2}$$
By continuity of $2^t$ making the sutbtituion $x=2^t$ and as $x\rightarrow \infty$ then.$t\rightarrow \infty$
Now $(2)$ is changed to $$\lim_{t\rightarrow \infty}{\frac{t}{2^t}}=0$$
A: \begin{gather*}
|x|<1:\quad f(x)=\sum_{n=1}^{\infty} x^n=\frac{x}{1-x} \\
xf'(x)=\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}
\end{gather*}
Let $x=\frac{1}{2}$
A: Since infinite series with nonnegative terms can be rearranged arbitrarily,
$$\sum_{i=1}^\infty \frac{i}{2^i} =
\sum_{i=1}^\infty \sum_{j=1}^i \frac{1}{2^i}
= \sum_{j=1}^\infty \sum_{i=j}^\infty \frac{1}{2^i}
= \sum_{j=1}^\infty \frac{1}{2^{j-1}}
= 2
$$
More graphically,
  1/2 + 2/4 + 3/8 + 4/16 + ...

= 1/2 + 1/4 + 1/8 + 1/16 + ...    (= 1)
      + 1/4 + 1/8 + 1/16 + ...    (= 1/2)           
            + 1/8 + 1/16 + ...    (= 1/4)
                  + 1/16 + ...    (= 1/8)
                        ....       ....

A: Start with the geometric series
$$s(x) = \sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$$
Then
$$x \frac{d}{dx} s(x) = \sum_{k=1}^{\infty} k x^k = \frac{x}{(1-x)^2}$$
Your case has $x=1/2$.
A: Let $s = 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + \dots$
Then $2s = 1 + 2/2 + 3/4 + 4/8 + 5/16 + \dots$
And then subtracting terms with similar denominators gives: $2s - s = 1  +1/2+1/4+1/8+1/16+\dots = 2$
A: Try to show that: 
$$\sum_{i=1}^{n}\frac{i}{2^i} = 2 - \frac{n+2}{2^n}$$
You can do this by induction. 
Than the step $n\rightarrow \infty$ will be easy. 
A: to see visually what the other answerers wrote, just explode the terms of the sum in this way:
1/2
1/4  1/4
1/8  1/8  1/8
1/16 1/16 1/16 1/16
1/32 1/32 1/32 1/32 1/32
...  ...  ...  ...  ...

Since all elements are positive, you may sum them in the order you want. Start summing each column, and obtain
1    1/2  1/4  1/8  1/16 ...

and now you're done.
A: Here is a visual proof that $$\frac14+\frac28+\frac3{16}+\frac{4}{32}+\frac{5}{64}+\cdots=1$$ which is the same relation after dividing by $2$ on each side:

If it is not clear how, here some areas are labeled (although I prefer the aesthetics of the unlabeled version). The rectangles fill up this $1\times1$ square, and for $i\in\mathbb{N}$, there are $i$ rectangles of area $\frac{1}{2^{n+1}}$.

A: A bit late.. but all the calculus and double sums aren't necessary.
$$\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^n}=\sum_{n=0}^{\infty}\frac{n+1}{2^{n+1}}=\sum_{n=1}^{\infty}\frac{n}{2^{n+1}}+\sum_{n=0}^\infty\frac{1}{2^{n+1}}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{n}{2^{n}}+1$$
Hence $$\displaystyle \sum_{n=1}^{\infty}\frac{n}{2^{n}}=2$$
A: It's fairly simple. Look at it this way:
$$
\sum_{i=1}^\infty \frac{i}{2^i} = \sum_{i=1}^\infty \frac{\sum_{k=1}^i 1}{2^i} = \sum_{i=1}^\infty \sum_{k=1}^i \frac{1}{2^i}
$$
From here, we just change the order of addition. Rather than adding along $k$, and then $i$, we add along $j=i-k$, and then along $k$. This turns our double sum into
$$
\sum_{i=1}^\infty \frac{i}{2^i} = \sum_{k=1}^\infty \sum_{j=0}^\infty \frac{1}{2^{j+k}} = \sum_{k=1}^\infty \frac{1}{2^k} \sum_{j=0}^\infty \frac{1}{2^j} = 1\cdot 2 = 2
$$
A: Consider the function
$$
f(z) = \sum_{i=0}^{\infty}\frac{z^i}{2^i} = \frac z{2-z}.
$$
Its radius of convergence is $2$.
Its derivative is
$$
f'(z) = \sum_{i=1}^{\infty}\frac{i}{2^i}z^{i-1} = \frac 2{(2-z)^2}.
$$
Since $1<2$, it is allowed to evaluate the derivative at $z=1$ with the written formula and obtain
$$
\sum_{i=1}^{\infty}\frac{i}{2^i} = f'(1) = \frac 2{(2-1)^2} = 2.
$$
A: the sequence $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+.....$$
is an $arithmetico  geometric series$
the sum of an$AG$ series of the form
$$S_{\infty}=a+(a+d)r+(a+2d)r^2+(a+3d)r^3+.....\infty=\frac{a}{1-r}+\frac{dr}{(1-r)^2}$$converting the above series inti this form
$$\frac{1}{2}( \frac{1}{1} +\frac{2}{2}+ \frac{3}{4}+...... )$$
here$a=1$,$d=1$,$r=\frac{1}{2}$.
you get the answer $2$.
