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$$\frac{1}{2}+ \frac{1}{2^2}+\ldots+\frac{1}{2^n} = 1 - \frac{1}{2^n}$$ I can do induction but I'm really struggling on this one in the proof part. It's the fractions and exponents that are throwing me off. Induction problem

plop
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toby843
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  • Please don't use pictures. – Dietrich Burde Jun 14 '21 at 14:58
  • Hi, please type out the math in the question using MathJax. Don't post pictures and show us your work or explain where you are stuck. – Stefan Octavian Jun 14 '21 at 15:00
  • For induction: [1] How much is $\displaystyle \left[1 - \frac{1}{2^n}\right] + \frac{1}{2^{n+1}}$? and [2] What is the relevance of the question that I posed in [1]? – user2661923 Jun 14 '21 at 15:01
  • Hi, sorry everyone for not knowing the regs on pictures etc. So I work through my base case, then assume p(k) is true and then try to prove p(k+1). So my work looks like this: – toby843 Jun 14 '21 at 15:05
  • Recommended reading: How to write a clear induction proof? – JMoravitz Jun 14 '21 at 15:06
  • my work looks like this when I start the proof: Hi, sorry everyone for not knowing the regs on pictures etc. So I work through my base case, then assume p(k) is true and then try to prove p(k+1). So my work as I start the proof looks like this: 1 - \frac{1}{2^k} + \frac{1}{2^k+1)} I then am struggling to work out how to simplify that to my end goal of 1- \frac{1}{2^k+1} – toby843 Jun 14 '21 at 15:15
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    I'm struggling to learn mathjax sorry – toby843 Jun 14 '21 at 15:15
  • Do you know the formula for the sum of consecutive terms in a geometric sequence? – Bernard Jun 14 '21 at 15:19
  • Hi, thanks for the induction proof guide. I don't think it's the induction process that's stumping me. It's the actual algebra involved. I'm sorry but I've come late to maths and am trying to learn as I go. I'm not sure what 1 - 1/2^n would be, it depends on n and I don't know how to vocalise it. If n = 2 it'd be 3/4. This is what's confusing me. I'm always left with something on top of the 1/2^n+1 that I need and I can't see how to get rid of it. – toby843 Jun 14 '21 at 15:23
  • sn = a1(1-r^n)/1-r – toby843 Jun 14 '21 at 15:23
  • Hint: If you absolutely have to use induction: $;\dfrac1{2^n}=\dfrac 2{2^{n+1}}$. – Bernard Jun 14 '21 at 15:29
  • Does this answer your question? How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$ – Sil Jun 14 '21 at 15:45
  • "I'm not sure what 1 - 1/2^n would be" It doesn't matter but it would be $\frac {2^n -1}{2^n}$. You can do the induction either by noting $(1-\frac 1{2^n}) + \frac 1{2^{n+1}} = 1 +(\frac 1{2^{n+1}} - \frac 1{2^n}) = 1-\frac 1{2^{n+1}}$ or by noting that $(1-\frac 1{2^n})+\frac 1{2^{n+1}} = \frac {2^n -1}{2^n} + \frac 1{2^{n+1}}= \frac {2^{n+1}-2}{2^{n+1}} + \frac 1{2^{n+1}} = \frac {2^{n+1} -1}{2^{n+1}} = 1 - \frac 1{2^{n+1}}$. – fleablood Jun 14 '21 at 16:04
  • "Do you know the formula for the sum of consecutive terms in a geometric sequence?" Not useful as that is exactly what is being asked to be proven. – fleablood Jun 14 '21 at 16:06
  • " 1 - \frac{1}{2^k} + \frac{1}{2^k+1)} I then am struggling to work out how to simplify that to my end goal of 1- \frac{1}{2^k+1} " $$1 -\frac 1{2^k} +\frac 1{2^{k+1}} = 1+(\frac 1{2^{k+1}} - \frac 1{2^k}) = 1 + (\frac 1{2^{k+1}} - \frac 1{2^k}\cdot \frac 22) = 1+(\frac 1{2^{k+1}} - \frac 2{2^{k+1}}) =1+(\frac {1-2}{2^{k+1}}) = 1 + (\frac{-1}{2^{k+1}}) = 1 - \frac 1{2^{k+1}}$$ – fleablood Jun 14 '21 at 16:10
  • .. or in other words.... put $-\frac 1{2^n}$ and $\frac 1{2^{k+1}}$ over a common denominator and combine. – fleablood Jun 14 '21 at 16:13

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It's obvious for $n=1$, so suppose its true for some $n\in\mathbb{N}$, i.e., $$\frac{1}{2}+\frac{1}{2^2}+\dots+\frac{1}{2^n}=1-\frac{1}{2^n}$$ We have $$\begin{align*} \left(\frac{1}{2}+\frac{1}{2^2}+\dots+\frac{1}{2^n}\right)+\frac{1}{2^{n+1}} &=1-\frac{1}{2^n}+\frac{1}{2^{n+1}}\\ &=1+\frac{1}{2^{n+1}}-\frac{2}{2^{n+1}}\\ &=1+\frac{1-2}{2^{n+1}}\\ &=1-\frac{1}{2^{n+1}}\end{align*}$$ which completes the induction.

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