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So I'm having trouble with proving this homework question by induction. $$ \frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{n-1}{2^{n-1}} + \frac{n}{2^n} <2 $$ I know how to prove that the series converges to 2 (using things like the ratio method), but actually using induction is where I get confused.

Base case is easy, n=1. $$ \frac{1}{2^1}<2 $$

Induction case we assume that $$ \frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} <2 $$

Then we get to fun old induction. How do I show that

$$ \frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} + \frac{k+1}{2^{k+1}} <2 ? $$

Sammy Aranny
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  • I think you have to consider $k < 2^{k+1}$ instead by multiplying by $2^k$ in both sides. – PackSciences Sep 12 '18 at 21:05
  • A different approach would be that $\sum_{n=0}^\infty nx^n = \frac{x}{(1-x)^2}$, for $x=1/2$ we have $2$. This also tells us that $2$ is the lowest possible bound. – Jakobian Sep 12 '18 at 21:10

5 Answers5

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Hint: $$ \begin{align} \frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} + \frac{k+1}{2^{k+1}} &=\qquad\;\;\frac12\Big(\frac{1}{2^1} + \frac{2}{2^2} + \dots +\frac{k-1}{2^{k-1}} + \frac{k}{2^k}\Big) \\ &\quad+\Big(\frac{1}{2^1} + \frac{1}{2^2}+\frac1{2^3} + \dots + \;\;\frac{1}{2^k}\;\;+\frac1{2^{k+1}}\Big) \end{align} $$ On the right hand side, the first summand corresponds to the induction hypothesis, and you can bound the second summand by...

Mike Earnest
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  • Working off the two summands you have, I can find an answer to the question.

    But I'm not entirely sure how you split up the equation like that.

     – Sammy Aranny Sep 12 '18 at 21:20
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    @SammyAranny I am confident you can figure that part out. – Mike Earnest Sep 12 '18 at 21:25
  • Took me a while but I figured it out, thank you so much!! – Sammy Aranny Sep 13 '18 at 02:09
  • Note that also that is a homwork! – user Sep 27 '18 at 14:43
  • @gimusi The reward to the asker should match the effort of the asker. Else we encourage people posting to MSE before thinking about the problem at all, which is not beneficial to their learning. Your answer rewarded an asker who gave no effort. The purpose of my downvote was to discourage this behavior. My answer here was to an asker which provided context and showed effort. – Mike Earnest Sep 27 '18 at 14:46
  • @MikeEarnest I gave a simple hint, it can look like a full solutionto you but for a not skilled person with this kind a doubts it require much work to find the final solution. Pay attention to ask good behaviours if you can't give the good example! Bye – user Sep 27 '18 at 14:50
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You can try to do the following: write

$$ \frac{1}{2} + \frac{2}{2^2} + \cdots + \frac{k}{2^k} + \frac{k+1}{2^{k+1}} = $$ $$ \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^k} + \frac{1}{2^{k+1}} + $$ $$ \frac{1}{2}\left(\frac{1}{2} + \frac{2}{2^2} + \cdots + \frac{k}{2^k}\right).$$

Use that the first summand is bounded by $1$, and, by induction hypothesis, the second one is strictly less than $\frac{1}{2} \cdot 2.$ This finishes the inductive step.

João Ramos
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  • Thank you, but how did you manage to split up the equation into two summands? – Sammy Aranny Sep 12 '18 at 21:23
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    Well, you need to break it up in a way that enables you to use induction step. In fact, you would like to write it as some function of the previous term. From that point on you manage to guess breaking up the way I did works. If the question is about 'why' this identity holds, I leave the justification for you - try 'regrouping' terms! – João Ramos Sep 12 '18 at 21:42
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Denote $$S_n=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}.\tag1$$Then $$\frac{1}{2}S_n=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\cdots+\frac{n}{2^{n+1}}.\tag2$$Thus by $(1)-(2)$, we obtain $$\frac{1}{2}S_n=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^n}\right)-\frac{n}{2^{n+1}}< 1-\frac{n}{2^{n+1}}<1,$$ which implies $$S_n < 2.$$

mengdie1982
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Hint:

It's a sum of an arithmetico-geometric progression:

$$S_{n}={\frac {a-(a+nd)\,r^{n}}{1-r}}+{\frac {dr\,(1-r^{n})}{(1-r)^{2}}}$$

Here $d$ is the common difference, $r$ is the common ratio, $n$ denotes the number of terms and $a$ is the first term.

paulplusx
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In that case a possible trick to use induction is to prove the stronger condition

$$\frac{1}{2} + \frac{2}{2^2} + \ldots +\frac{n-1}{2^{n-1}} + \frac{n}{2^{n}} <2-\frac{n+1}{2^{n-1}}<2$$

and the induction step becomes

$$\frac{1}{2} + \frac{2}{2^2} + \ldots + \frac{n}{2^n} + \frac{n+1}{2^{n+1}}\stackrel{Ind. Hyp.}<2-\frac{n+1}{2^{n-1}}+ \frac{n+1}{2^{n+1}}\stackrel{?}<2-\frac{n+2}{2^{n}}$$

and the last inequality holds indeed

$$2-\frac{n+1}{2^{n-1}}+ \frac{n+1}{2^{n+1}}\stackrel{?}<2-\frac{n+2}{2^{n}}$$

$$\frac{n+1}{2^{n-1}}- \frac{n+1}{2^{n+1}}\stackrel{?}>\frac{n+2}{2^{n}}$$

$$4(n+1)-(n+1)\stackrel{?}> 2(n+2)$$

$$3n+3\stackrel{?}>2n+4$$

wich is true for $n>1$.

user
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