Recurring Recursion

Algebra Level 3

a n = ∏ k = 1 n − 1 a k ∑ k = 1 n − 1 a k {a_n} = \frac{\displaystyle \prod_{k=1}^{n-1} a_k}{\displaystyle \sum_{k=1}^{n-1} a_k}

If the sequence a n a_n satisfy the recurrence relations formula as described above with a 1 = 1 a_1 = 1 . Determine the value of ∑ i = 1 7 1 a i \displaystyle \sum_{i=1}^7 \frac1{a_i} .

You might want to use a calculator as the values increase rapidly.


The answer is 547166.

Efren Medallo by Efren Medallo , 26, Philippines

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1 solution

Efren Medallo
Jul 2, 2015

This is the crudest solution. I would like to give the opportunity to you guys to do an alternative one.

The first two numbers of the sequence would both be one. I think that can easily be seen.

for a 3 a_{3} , and further, it would entail direct substitution.

That is, a 3 = 1 ⋅ 1 1 + 1 = 1 2 a_{3} = \large \frac {1 \cdot 1 }{1 + 1 } = \frac {1}{2} .

then, a 4 = 1 ⋅ 1 ⋅ 1 2 1 + 1 + 1 2 = 1 5 a_{4} = \large \frac {1 \cdot 1 \cdot \frac{1}{2} }{1 + 1 + \frac{1}{2} } = \frac {1}{5} .

then, a 5 = 1 ⋅ 1 ⋅ 1 2 ⋅ 1 5 1 + 1 + 1 2 + 1 5 = 1 27 a_{5} = \large \frac {1 \cdot 1 \cdot \frac{1}{2} \cdot \frac{1}{5} }{1 + 1 + \frac{1}{2} + \frac{1}{5}} = \frac {1}{27} .

then, a 6 = 1 ⋅ 1 ⋅ 1 2 ⋅ 1 5 ⋅ 1 27 1 + 1 + 1 2 + 1 5 + 1 27 = 1 739 a_{6} = \large \frac {1 \cdot 1 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot \frac {1}{27} }{1 + 1 + \frac{1}{2} + \frac{1}{5} + \frac {1}{27}} = \frac {1}{739} .

then a 7 = 1 ⋅ 1 ⋅ 1 2 ⋅ 1 5 ⋅ 1 27 ⋅ 1 739 1 + 1 + 1 2 + 1 5 + 1 27 + 1 739 = 1 546391 a_{7} = \large \frac {1 \cdot 1 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot \frac {1}{27} \cdot \frac {1}{739} }{1 + 1 + \frac{1}{2} + \frac{1}{5} + \frac {1}{27} + \frac {1}{739} } = \frac {1}{546391} .

And so, 1 + 1 + 2 + 5 + 27 + 739 + 546391 = 547166 1 + 1 + 2 + 5 + 27 + 739 + 546391 = \boxed{547166} . This will just be a reference to check whether your solutions are correct. :) Please don't forget to reshare this problem!

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What is the general term like? Or is this a purely computational problem? (which is why we don't think it's level 5)

Does the limit of ∑ a i \sum a_i exist?

Calvin Lin Staff - 5 years, 11 months ago

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Nope. A general term exists.

a n = 1 2 ∏ k = 1 n − 1 a k + ∑ j = 1 n − 1 ∏ k = 1 n − 1 a k a j a_{n} = \large \frac {1}{ 2\prod_{k=1}^{n-1} a_{k} + \sum_{j=1}^{n-1} \frac{\prod_{k=1}^{n-1} a_{k}}{a_{j}} } .

Efren Medallo - 5 years, 11 months ago

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And yes, a limit exists for such a summation.

Efren Medallo - 5 years, 11 months ago

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@Efren Medallo – Is there a closed form for the general term? IE that it doesn't depend on the previous a i a_i .

What is the value of the limit, and why? Can it be calculated theoretically, instead of "we find the sum of the first few terms, which gives us a close enough value"?

Calvin Lin Staff - 5 years, 11 months ago

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