6
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I'm in trouble with this homework.

Find the sum of the series $ 1 + \frac12 + \frac13 + \frac14 + \frac16 + \frac18 + \frac19 + \frac1{12} +\cdots$, where the terms are the inverse of the positive integers whose only prime factors are 2 and 3.

Hint: write the series as a product of two geometric series.

Ok, I found some patterns in these numbers, but I can't find the two series.

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11
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$$\left(1+\frac{1}{2}+\frac{1}{4}+\cdots \right)\left(1+\frac{1}{3}+\frac{1}{9}+\cdots \right)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\cdots$$

both of the first series are geometric and can be summed $$1+\frac{1}{2}+\frac{1}{4}+\cdots =2$$ $$1+\frac{1}{3}+\frac{1}{9}+\cdots =\frac{3}{2}$$ so $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\cdots=3$$

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14
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Hint: $$ = \left( 1+\frac{1}2 + \frac{1}4 + \cdots \right) \left( 1+ \frac{1}3 + \frac{1}9 + \cdots \right)$$.

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  • $\begingroup$ Ok, I supposed that you mean $ \frac1{2^n} $ and $ \frac1{3^n} $, but what about $ \frac1{12} $? $\endgroup$ – Luciano Gerhardt Jun 9 '14 at 0:11
  • $\begingroup$ The comment was a mistake, I don't think the Cauchy product will help understand this. Just try and multiply $\sum_{i=0}^n\frac{1}{2^i}$ with $\sum_{k=0}^n\frac{1}{3^k}$ for small $n$ and you'll see what's happening. Basically, all you're doing is using distributivity. For a formal proof, you'd also have to mention that the two series into which you decompose your series are both absolutely convergent, and so any reordering of the two series (and of their product) is also absolutely convergent. $\endgroup$ – G. Bach Jun 9 '14 at 0:23
  • $\begingroup$ I think I understand, just needed to be sure! Thank you! $\endgroup$ – Luciano Gerhardt Jun 9 '14 at 0:28
  • $\begingroup$ You could refer to the Cauchy product like so: $1 + \frac12 + \frac13 + \frac14 + \frac16 + \frac18 + \frac19 + \frac1{12} +\ldots = \underbrace{(1)}_{c_0} + \underbrace{\left(\frac12 + \frac13\right)}_{c_1} + \underbrace{\left(\frac14 + \frac16 + \frac19\right)}_{c_2} + \frac18 + \ldots =$ $\left(\sum_{i=0}^\infty\frac{1}{2^i}\right) \star \left(\sum_{i=0}^\infty\frac{1}{3^i}\right)$ where $\star$ is the Cauchy product of two series. You'll also need to reorder the original sum, though. $\endgroup$ – G. Bach Jun 9 '14 at 0:31

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