<s>[INST] Induction proof: $\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}$ $<2$ Prove by induction the following.
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}<2.$$
Caveat: The $<$ will be hard to work with directly. Instead, the equation above can be written in the form,
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}=2-\text{something}.$$
First, find the "something" and then use that form of the equation to prove the assertion.
I can't seem to figure out the form that this equation can be written as. Also, once I find the form how would I do the proof.
I understand it involves using a Basic Step and an Induction Step
 [/INST] This question is circling in rounds.
Because, as everybody knows, the sum tends to $2$ so that in 
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}+ r_n=2,$$ the expression of $r_n$ must be exact!
If you try with an inequality such as
$$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +...+\frac{n}{2^n}+ r_n<2,$$
induction will require $r_n\ge\dfrac{n+1}{2^{n+1}}$ to absorb the next term, but if $r_n$ is not tight sooner or later the sum with the remainder will exceed $2$.

Robert Z. found a nice workaround.
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