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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

Robert Z
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Viserom
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  • Can you see how to relate your sum to a derivative ? – Maxime Ramzi Oct 19 '18 at 06:49
  • See here (with $r=1/2$): https://math.stackexchange.com/questions/119636/formula-for-calculating-sum-n-0mnrn – Hans Lundmark Oct 19 '18 at 06:54
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    IMO, this is a poor question because it virtually requires that you solve the problem non inductively before you can start the inductive proof. –  Oct 19 '18 at 07:10

3 Answers3

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Show that $$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}\leq 2-\frac{An+B}{2^n}$$ for some $A,B\geq 0$. Then the base case is satisfied if $$\frac{1}{2}\leq 2-\frac{A+B}{2}$$ that is $A+B\leq 3$. The induction step works if for all $n\geq 1$, $$2-\frac{An+B}{2^n}+\frac{n+1}{2^{n+1}}\leq2-\frac{A(n+1)+B}{2^{n+1}},$$ that is $$n+1\leq (2An+2B)-(A(n+1)+B)=An+B-A.$$ It follows that $A=1$ and $B=2$ and we may conclude that:

For all $n\geq 1$, $$\frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} +\dots+\frac{n}{2^n}\leq 2-\frac{n+2}{2^n}$$

P.S. Looking back to the the induction proof steps it is easy to realize that the above inequality is actually an equality!!

Robert Z
  • 145,942
  • Well done, using an ansatz is indeed a good way. (I would have added an $n^2$ term just to see, because the numerators form a triangular number ;-) ) –  Oct 19 '18 at 07:29
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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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Let the sum of the $n$ first terms be $S_n$. We have the recurrence

$$S_{n+1}=S_n+\frac{n+1}{2^{n+1}}$$ or

$$2^{n+1}S_{n+1}=2\cdot2^nS_n+n+1.$$

This hints the change of variable that leads to

$$R_{n+1}=2R_n+n+1.$$

This is a linear recurrence which we will solve a usual:

  • homogeneous part, $R_{n+1}=2R_n$, so that $R_n=R_12^{n-1}$.

  • particular solution found by indeterminate coefficients, using a linear ansatz: $-n-2$.

Now, using the initial condition $R_1=1$,

$$S_n=2-\frac{n+2}{2^n}.$$

We can complete the inductive proof:

$$S_1=\frac12=2-\frac{1+2}{2^1}<2,$$

$$S_n=2-\frac{n+2}{2^n}<2\implies S_{n+1}=2-\frac{n+2}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{(n+1)+2}{2^{n+1}}<2.$$