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I'm trying to study for a test and one of the practise questions is very confusing and not sure what to do:

Prove by induction that $$\sum_{i=1}^n \frac{3}{4^i} < 1$$ for all $n \geq 2$

The furthest I'm able to get is getting rid of the summation.

So I get $(3/4)^i n< 1$ for all $n \geq2$.

Any help would be greatly appreciated. Thanks.

Brian M. Scott
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Abushawish
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  • You should take a look at http://math.stackexchange.com/questions/19485/dominoes-and-induction-or-how-does-induction-work/19488#19488 – Ross Millikan Mar 26 '13 at 01:34
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    I think the question is suppose to be $> 1$ not $< 1$. It's fairly straight-forward to show, by induction, however induction is not really necessary in this case. – Noble. Mar 26 '13 at 01:38
  • Please check that you have the question exactly right. For large $n$ your sum is close to $3$. – André Nicolas Mar 26 '13 at 01:40
  • Are you sure that it isn’t $$\sum_{i=1}^n\frac3{4^i}<1;?$$ If it is supposed to have the exponent only in the denominator, this is a duplicate of this question. – Brian M. Scott Mar 26 '13 at 01:55
  • Yeah it's less than one:

    http://i.imgur.com/MxEnMWm.png

     – Abushawish Mar 26 '13 at 01:57
  • You misunderstood the problem, which in fact is a duplicate of this question: $3/4^i$ is $\dfrac3{4^i}$, not $\left(\dfrac34\right)^i$. – Brian M. Scott Mar 26 '13 at 02:23
  • Ah, I'm the one at fault for the edit that made it $(3/4)^i$ in the sum. Sorry - I saw the brackets in the "So I get..." line, and assumed that leaving off the brackets in the sum was the mistake. – Glen O Mar 26 '13 at 07:07

3 Answers3

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It is defeating you because $${3\over 4} + {9\over 16} = {21\over 16} > 1$$

ncmathsadist
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Prove that it's true for $n=2$.

Then prove that, if it is true for $n=k$, then it will be true for $n=k+1$.

That's the inductive process.

The Chaz 2.0
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Assuming the question is supposed to be $\displaystyle\sum_{i=1}^n \left(\dfrac{3}{4}\right)^i > 1, \ \ \forall n \geq 2$ it's pretty straight-forward to do.

Clearly it's true for $n=2$ as we get $\displaystyle\sum_{i=1}^2 \left(\frac{3}{4}\right)^i = \frac{3}{4} + \frac{9}{16} = \frac{21}{16} > 1$

Assume then, for $n=k$ we have that $\displaystyle\sum_{i=1}^k \left(\frac{3}{4}\right)^i > 1$

Then for $n=k+1$

$\displaystyle\sum_{i=1}^{k+1} \left(\frac{3}{4}\right)^i = \displaystyle\sum_{i=1}^k \left(\frac{3}{4}\right)^i + \left(\frac{3}{4}\right)^k > 1 + \left(\frac{3}{4}\right)^k > 1 $

EDIT: I just thought I'd add something, I don't know if you just typed the question out incorrectly or if you really have been trying to show that the sum is $<1$, but it's very important to show it is at least true for one value of $n$ in the given range (so in this case any $n \geq 2$) because if you don't do this, the proof (by induction) doesn't really work. The power of showing it's true for $n=2$, means that once you've proven it by induction, it's also true for $n=3,4,5,6...$ because of the way in which you assume it's true for some $k$ and prove it's true for $k+1$.

Noble.
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