How to show $\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$ using induction I´d like to show that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}<2$$ using the fact that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$$
I guess the answer use transitivity of natural numbers and the inequality $$n+1<2^n$$ In this case, i can use it because i suppose it for natural n such that n>1. I'd like to have something like  $$....+(n+1)<2(2^{n+1})$$ because you can divide both sides by $2^{n+1}$, and so, you get $$...+\frac{n+1}{2^{n+1}}<2$$
I´ve noticed that $$2^n+n+1<2^n+2^n=2^{n+1}<2(2^{n+1})$$ and it leads me to think $$...+n+1<2^n+n+1$$ i´ve tried dividing both sides ,of the inequality i´m assuming as a fact, by $2^n/2$ in order to get $2^n$ 
in the right hand of the inequality and then, add $n+1$ to both sides: $$(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n})\frac{2^n}{2}+n+1<2(\frac{2^n}{2})+n+1=2^n+n+1$$ but at this part i get stuck because if i continue it doesn't lead me to what i´d like to show. 
Please, help me. I hope someone here have already proved this before, because i don't find anything similar to this at any part of the web.

In order to use induction in the way you tried, you may show a stronger inequality which implies the original one, such as
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}\leq 2-\frac{n+a}{2^n}$$
where $a$ is a non-negative constant to be determined.
The basic step: for $n=1$ is $\frac{1}{2}\leq 2-\frac{1+a}{2}$, that is $a\leq 2$.
Then, at the inductive step, for $n\geq 1$,
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}\leq \left(2-\frac{n+a}{2^n}\right)+\frac{n+1}{2^{n+1}}\stackrel{?}{\leq}2-\frac{n+1+a}{2^{n+1}}$$
and the last inequality simplifies to $a\geq 2$.
So $a=2$ works!! 
