Summing Specific Reciprocals

Algebra Level 3

$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{12}+\dfrac{1}{16}+\dfrac{1}{18}+\cdots = \, ?$

Clarification : The denominators are all positive integers whose prime divisors are either 2 and/or 3 only. In other words, they exclude all numbers with a prime factor greater than 3.

The answer is 3.0.

4 solutions

Julian Yu
Aug 11, 2016

Relevant wiki: Geometric Progression Sum

The sum can be represented by the product $\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \right) \left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots \right)$

$=\left(\frac{1}{1-\frac{1}{2}}\right)\left(\frac{1}{1-\frac{1}{3}}\right)$

$=3.\square$

Moderator note:

Correct! The penultimate step follows from the infinite geometric progression sum

$a + ar + ar^2 + ar^3 + \cdots = \dfrac a{1-r}, \quad |r| < 1 .$

Bonus : Can you generalize this? That is, what is the sum of the reciprocals of positive integers whose prime factors are at least of the numbers $2,3,5, \ldots, p_{n}$ only?

This is actually a pretty elegant solution. Well done! It's the same as evaluating

$\displaystyle \sum _{ a=0 }^{ \infty }{\displaystyle \sum _{ b=0 }^{ \infty }{ \dfrac { 1 }{ { 2 }^{ a }{ 3 }^{ b } } } } =3$

The key is recognizing that each denominator is an unique product of powers of $2$ and $3$ , so that your product of infinite series generates only such unique denominators.

Michael Mendrin - 4 years, 10 months ago

I see...that it why i didnt go to law school. I missed the line prime divisor greater than 2 or 3

Greg Grapsas - 4 years, 10 months ago

I liked the problem and the solution. Thanks so much for sharing such an awesome problem. :)

However, you can make your solution more detailed by mentioning the sum of infinite geometric progression formulas.

Sandeep Bhardwaj - 4 years, 10 months ago

This is the way I tried to approach the problem. Since the numbers are divisible by 2 or 3 only. So ,

$1 + \frac{1}{2} [ 1 + \frac{1}{2} +\frac{1}{3}.....] + \frac{1}{3}[ 1 + \frac{1}{2} +\frac{1}{3}.....] - \frac{2}{6} [ 1 + \frac{1}{2} +\frac{1}{3}.....] ..$

$= 1 + \frac{1}{2} [ 1 + \frac{1}{2} +\frac{1}{3}.....]$

And I searched for the value of $[ 1 + \frac{1}{2} +\frac{1}{3}.....]$ And they tell it will never converge and so will tend to infinity. So how could the answer come 3 . Please explain.

Anurag Pandey - 4 years, 10 months ago

Please add in $\frac18$

Shaun Leong - 4 years, 10 months ago

Yeah, added. Thanks :)

Sandeep Bhardwaj - 4 years, 10 months ago

I didn't got ya? Could you please explain how this product will give the same value of the given series?

Anurag Pandey - 4 years, 10 months ago

Notice the similarity with the Euler Product

Michael Mendrin - 4 years, 10 months ago

So can you explain where the 1/6 gone

Konuralp Şenoğlu - 4 years, 10 months ago

Nvm got it

Konuralp Şenoğlu - 4 years, 10 months ago

It does not converge to 3.33333 nor 3. Thru 1/300 already has a sum around 4.77+

Greg Grapsas - 4 years, 10 months ago

It does converge. Check the partial sums again.

$\sum _{ x=0 }^{ \infty }{ \sum _{ y=0 }^{ \infty }{ \frac { 1 }{ { 2 }^{ x }{ 3 }^{ y } } } } =\left( \sum _{ x=0 }^{ \infty }{ \frac { 1 }{ { 2 }^{ x } } } \right) \left( \sum _{ y=0 }^{ \infty }{ \frac { 1 }{ { 3 }^{ y } } } \right)$

Arulx Z - 4 years, 10 months ago

I dont dispute convergence...but the number is not in the 3s.

Greg Grapsas - 4 years, 10 months ago

The problem is not restricte to powers of 2 or 3....it says divisors of 2 or 3. 1/21 is an element of the series and it does not represent a product of a power of 2 and or 3

Greg Grapsas - 4 years, 10 months ago

I agree that it is a bit unclear but the problem states that

In other words, it excludes any positive integer that is divisible by a prime greater than 3.

Arulx Z - 4 years, 10 months ago

what a champion

Grant Bulaong - 4 years, 10 months ago

Bonus 2: Show that the value of the sum equals to $\dfrac{n}{\phi\left(n\right)}$ (Euler's Totient Function) if we have a finite number of primes which have a product of $n$ instead of only $2$ and $3$ .

Bonus 3: Show that $1 + \dfrac12 + \dfrac13 + \cdots$ does not have a finite value.

Jesse Nieminen - 4 years, 9 months ago

The problem says "In other words, they exclude all the prime numbers greater than 3." This sort of implies that it does not exclude anything but the primes greater than 3, so, say "1/35" would be in the series. Of course the problem is still accurate but I feel it would be confused.

Alex Li - 4 years, 4 months ago

Fixed, thanks

Julian Yu - 4 years, 4 months ago

Chew-Seong Cheong
Aug 11, 2016

$\begin{aligned} S & = \sum_{j=0}^\infty \sum_{k=0}^\infty \frac 1{2^j 3^k} \\ & = \sum_{j=0}^\infty \frac 1{2^j} \sum_{k=0}^\infty \frac 1{3^k} \\ & = \frac 1{1-\frac 12} \cdot \frac 1{1-\frac 13} \\ & = 2 \cdot \frac 32 = \boxed{3} \end{aligned}$

Eilon Reisin-Tzur
Aug 12, 2016

Let S be the original sum. Then S can be split into four sums: 1 + (1/2 + 1/4 + 1/8 + ...) + (1/3 + 1/9 + 1/27 + ...) + 1/6 (1 + 1/2 + 1/3 + 1/4 + 1/6 + ...) = 1 + 1 + 1/2 + 1/6 S = 5/2 + 1/6 S.

So S = 5/2 + 1/6 S and so S = 3.

Cool solution. I started off solving it this way but didn't get to the end. Ended up solving it exactly the way Chew-Seong Cheong did above.

Richard Costen - 4 years, 10 months ago

Razzi Masroor
Aug 18, 2016

For people who love geometric sequences, thats me. I didi it by separating it into 1 +1/2+1/4...... and 1/3+1/6+1/12.... and 1/9 +1/18+1/36.... 1+1/2+1/4....=2. and the infinite other sequences and 1/3 of each other.A Geometric sequence on a Geometric sequence. The a for a+an+an^2.... is equal to 1/3+1/6+1/12.... which is 2/3 while n is 1/3. Plugging it in the equation gets you 1. 2+1=3

Summing Specific Reciprocals

The answer is 3.0.

4 solutions

Moderator note:

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