Inequality Induction Proof, how should I proceed? I have been trying to prove inequalities using induction to no avail.
For example,
Prove the following:
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{n}{2^{n}}<2$
Base Case:
$\frac{1}{2^{1}} = 2$
$\frac{1}{2}<2\;$  Which is true.
We assume, for n=k, that
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{k}{2^{k}}<2\;\;\;$ is true.
As induction step with n=k+1, we need to prove that
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{k+1}{2^{k+1}}<2\;\;\;$
How should I proceed from here?
 A: A case can be made that this response is defective, because I am abandoning the OP's approach, which I consider the wrong approach for this problem.  This is a classic geometric series problem where (for $0 < t < 1$) you have
$$S = t + t^2 + t^3 + \cdots = t \times \frac{1}{1-t}$$
and
$$T = t + 2t^2 + 3t^3 + \cdots =
S(1 + t + t^2 + \cdots) =
t \times \left(\frac{1}{1-t}\right)^2.$$
A: You may first prove the following equality:

Prove that $\frac 1 {2^1} + \frac 2 {2^2} + \dots + \frac k {2^k} = 2 - \frac {k + 2} {2^k}$.

This can be proved by induction on $k$ (easy exercise).
Once it is proved, the original inequality follows.
A: I finally managed to prove it with induction, which was the goal.
I followed the suggestion to prove that  $\;\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+...+\frac{k}{2^{k}} = 2-\frac{k+2}{2^{k}}\;$
for k+1 we have that
$\;\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+...+\frac{k}{2^{k}}+\frac{k+1}{2^{k+1}} =   2-\frac{k+3}{2^{k+1}}\;$
It follows from the inductive hypothesis that
$2-\frac{k+2}{2^{k}}+\frac{k+1}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$
$2-\frac{2k+4}{2^{k+1}}+\frac{k+1}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$
$2+\frac{-3-k}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$
$2-\frac{k+3}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$
Which verifies that $\;\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+...+\frac{k}{2^{k}}$ is 2 minus something else, which in turn means that it is less than 2 thus proving the original statement.
