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 $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$

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 $$\lim_{t\rightarrow \infty}{\frac{t}{2^t}}=0$$
