<s>[INST] Why can't I prove this statement by simple induction? Sum of $1/2^1 + \cdots+ n/2^n = x$ I have to prove the following:
$$
\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{2 + n}{2^n}.
$$
I am trying to prove this by simple induction. First, I proved that $P(1)$ holds. It clearly does.
I then assume that $n$ is a positive number $> 1$ and that $P(n)$ holds. Hence,
$$
\frac{1}{2}+\ldots+\frac{n}{2^n}=2-\frac{2 + n}{2^n}.
$$
I now add ${\displaystyle \frac{n + 1}{2\cdot 2^n}}$ to both sides to get the following:
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2\cdot 2^n}=2-\frac{2+n}{2^n}+\frac{n+1}{2\cdot 2^n}.
$$
But after manipulating the right side I get the following:
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{3n + 5}{2^{n+1}}.
$$
This is definitely not right since I should be getting
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{n + 3}{2^{n+1}}=2-\frac{2+(n+1)}{2^{n+1}}
$$
to prove that $P(n)$ implies $P(n+1)$.
What am I doing wrong?
 [/INST] $$\begin{array}{l}
 \left| x \right| < 1:\frac{1}{{1 - x}} = \sum\limits_{k = 0}^\infty  {x^k }  = 1 + \sum\limits_{k = 1}^\infty  {x^k }  \\ 
  \Rightarrow \sum\limits_{k = 1}^\infty  {kx^k }  + \sum\limits_{k = 1}^\infty  {x^k }  = \frac{x}{{\left( {1 - x} \right)^2 }} + \frac{x}{{1 - x}} \\ 
  \Rightarrow \sum\limits_{k = 1}^\infty  {\left( {k + 1} \right)x^k }  = \frac{x}{{\left( {1 - x} \right)^2 }} + \frac{x}{{1 - x}} \\ 
 \left| x \right| < 1;x = \frac{1}{2} \Rightarrow \sum\limits_{k = 1}^\infty  {\left( {k + 1} \right)\left( {\frac{1}{2}} \right)^k }  = \frac{{\frac{1}{2}}}{{\left( {1 - \frac{1}{2}} \right)^2 }} + \frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}} = 3 \\ 
 \end{array}$$
</s>