My Math Forum Infinite Sum of the Inverses of Pronic Numbers

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 March 3rd, 2015, 04:46 AM #1 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Infinite Sum of the Inverses of Pronic Numbers So, 1/2 + 1/6 + 1/12 + 1/20 + .... = ? Just doing my usual recreational number crunching, it would appear that the denominators will always manage to stay a little bigger than the numerators, which would mean that this series has a limit of 1. Am I correct in this? Does it have an exact value of 1? Or something else a little less than 1? I tried to google up some insight, but came up empty. I don't have much background in infinite series. So any help will be appreciated. I put this in number theory because (a) that is where I hang and (b) this ultimately relates to some number theory stuff related to the mysteries of the multiplication table I am working on. If it belongs elsewhere, let me know!
 March 3rd, 2015, 05:53 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 It's 1. For finitely many terms, n, we have $\displaystyle \sum_{i=1}^{k} \frac{1}{i(i+1)}$ for some k. Suppose k = 1. Then we have $\displaystyle \frac{1}{2} = \frac{k}{k+1}$ Suppose $\displaystyle \sum_{i=1}^{k} \frac{1}{i(i+1)} = \frac{k}{k+1}$ Then $$\sum_{i=1}^{k+1} \frac{1}{i(i+1)} \\= \frac{1}{(k+1)(k+2)}+\sum_{i=1}^{k} \frac{1}{i(i+1)} \\= \frac{1}{(k+1)(k+2)} + \frac{k}{k+1} \\= \frac{1}{(k+1)(k+2)} + \frac{k(k+2)}{(k+1)(k+2)} \\= \frac{(k+1)^2}{(k+1)(k+2)}\\= \frac{(k+1)}{(k+2)}$$ Therefore, by induction, $\displaystyle \sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$ and as $\displaystyle \lim_{n \to \infty} \frac{n}{n+1} = 1$, the sum is 1. Thanks from CRGreathouse, topsquark and v8archie
 March 3rd, 2015, 06:05 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms It's not hard to prove, by induction, that $$\sum_{n=1}^\infty\frac{1}{n(n+1)} = \sum_{n=2}^\infty(-1)^n\zeta(n)$$ which allows a fast computation in PARI/GP as Code: sumalt(n=2,zeta(n)*(-1)^n) which verifies to lots of digits that this is 1. A good way to prove these identities is by the telescoping method. An easier method is to trawl through the OEIS https://oeis.org/A002378 where it turns out that Gary W. Adamson computed a (strange) equivalent version of this identity in 2003. If you want to prove it yourself, the sledgehammer to end all sledgehammers is A=B. Edit: An even easier method is to look at the post above this one, which I hadn't seen when composing this post. Thanks from Hoempa and topsquark
 March 3rd, 2015, 08:25 AM #4 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Thanks, guys! Thanks from Hoempa
March 3rd, 2015, 10:06 AM   #5
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Quote:
 Originally Posted by Hoempa Therefore, by induction, $\displaystyle \sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$
Is the any particular reason for the inductive proof? Rather easier would be the decomposition by partial fractions $${1 \over n(n+1)} = {1 \over n} - {1 \over n+1}$$ which gives a telescoping series $$\sum_{k = 1}^n {1 \over k(k+1)} = \sum_{k = 1}^n {1 \over k} - {1 \over k+1} = 1 - {1 \over n+ 1}$$
which easily gives the desired limit.

March 3rd, 2015, 12:26 PM   #6
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Quote:
 Originally Posted by v8archie Is the any particular reason for the inductive proof? Rather easier would be the decomposition by partial fractions $${1 \over n(n+1)} = {1 \over n} - {1 \over n+1}$$ which gives a telescoping series $$\sum_{k = 1}^n {1 \over k(k+1)} = \sum_{k = 1}^n {1 \over k} - {1 \over k+1} = 1 - {1 \over n+ 1}$$ which easily gives the desired limit.
Personally, as proofs are concerned, I say the more, the merrier! Thanks for this latest!

March 3rd, 2015, 02:17 PM   #7
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Quote:
 Originally Posted by v8archie Is the any particular reason for the inductive proof?
Not really, I suspected maybe a telescoping sum could work but I couldn't figure it out before I saw how it could be done with induction.

Nice to see all these methods for this!

 Tags infinite, inverses, numbers, pronic, sum

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