<s>[INST] 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?
 [/INST] 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.
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