Prove $1/2^1 + 2/2^2 +....+ n/2^n < 2$ using induction This is another problem with induction that I'm sure requires some "thinking outside the box" which is something I cannot apply with my way of solving these problems.
I need to now how I can complete the induction step and I need a solution using induction.

Suppose 
$\sum_\limits{k=1}^{\infty} k r^k$ converges  (and it does if r<1)
$S - rS = \sum_\limits{k=1}^{\infty} (n+1 - n) r^k = \sum_\limits{k=1}^{\infty}  r^k\\
S = \frac {r}{(1-r)^2}$
$\sum_\limits{k=1}^{\infty} k (\frac 12)^k = \frac {\frac 12}{(1-\frac 12)^2} = 2$
$\sum_\limits{k=1}^{n} k (\frac 12)^k < \sum_\limits{k=1}^{\infty} k (\frac 12)^k$
But that is not a proof by induction, is it?
Rather than prove that 
$a_n = \frac 12 + \frac 24 \cdots + \frac {n}{2^n} < 2$ it will be easier if we try to prove something more restrictive.  
Proposition:
$a_n = \frac 12 + \frac 24 \cdots + \frac {n}{2^n} = 2 - \frac {n+2}{2^n}$
Proof by induction:
base case:
n = 1
$\frac 12  = 2 - \frac {3}{2}$
Inductive hypothesis:
Suppose $a_n = \frac 12 + \frac 24 \cdots + \frac {n}{2^n} = 2 - \frac {n+2}{2^n}$
we will need to show:
$a_{n+1} <2-\frac {n+3}{2^{n+1}}$
$a_{n+1} = a_n + \frac {n+1}{2^{n+1}}\\
a_{n+1} = 2 - \frac {n+2}{2^n} + \frac {n+1}{2^{n+1}}$
Based on the inductive hypothesis
$a_{n+1} = $$2 - \frac {2(n+2)-(n+1)}{2^{n+1}}\\
2 - \frac {n+3}{2^{n+1}}\\
$
QED
