Wikipedia:Reference desk/Archives/Mathematics/2018 February 25

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February 25

Do these converge to nice numbers?

Do 1⁄(2*3) + 1⁄(3*5) + 1⁄(5*7) + ...

and 1⁄(2*3) + 1⁄(5*7) + 1⁄(11*13) + ...

(where the numbers in the denominators are primes) converge to known numbers? That is, I have their approximate numerical values - do they converge to something concise? Bubba73 ^{You talkin' to me?} 00:52, 25 February 2018 (UTC)

Not that I know of, but the 1st one is OEIS sequence A210473, and the 2nd one is related to OEIS sequence A089581. Iffy★Chat -- 10:45, 25 February 2018 (UTC)

Thanks. Bubba73 ^{You talkin' to me?} 17:41, 25 February 2018 (UTC)

I don't know whether you tried it, but if you have good numerical values, you can check the Inverse Symbolic Calculator to obtain a guess what the number's exact value could be. —Kusma (t·c) 20:26, 25 February 2018 (UTC)

The first sum (let's denote it $S$) can be calculated in the following way:

$S=\sum \limits _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+3)}}-{\frac {1}{6}}$.

On the other hand:

$S'=\sum \limits _{n=-\infty }^{\infty }{\frac {1}{(2n+1)(2n+3)}}=2*\sum \limits _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+3)}}-1$

from where it is clear that

$S={\frac {S'+1}{2}}-{\frac {1}{6}}$.

The value of $S'$ can be calculated using the following integral

$S'={\frac {1}{2i}}\oint \limits _{C}{\frac {\cot(\pi z)}{(2z+1)(2z+3)}}dz$,

where the closed contour $C$ is chosen to go from $-\infty$ to $\infty$ below the real axis and back above it. The only poles of the function under the integral are those of cotangent. Therefore $S'=0$. The final result is $S=1/3$.

The second sum can be calculated in a similar way.

Ruslik_Zero 20:39, 26 February 2018 (UTC)

The denominators in the series are actually the primes. Bubba73 ^{You talkin' to me?} 20:47, 26 February 2018 (UTC)

... so the first sum is probably closer to 0.3 than to 0.33, but I can't prove it. Dbfirs 08:49, 27 February 2018 (UTC)

Also, I'm pretty sure you can avoid complex analysis and get S = 1/3 using telescoping sums. In case anyone cares

1⁄(1*3) + 1⁄(5*7) + 1⁄(9*11) + ... = π⁄8. — Preceding unsigned comment added by RDBury (talk • contribs) 10:51, 27 February 2018 (UTC)

@RDBury: Can you give a reference for that result? I’d like to add it to our article List of formulae involving π. Thanks. Loraof (talk) 17:07, 27 February 2018 (UTC)

See Leibniz formula for π#Convergence. --RDBury (talk) 10:53, 28 February 2018 (UTC)

Yes, numerically I got about 0.30109317 for the first one and about 0.21042575 for the second one. Bubba73 ^{You talkin' to me?} 15:40, 27 February 2018 (UTC)

Sorry, I missed that they are primes. But you should explain better how these sequences are constructed. Do they involve only prime pairs? Ruslik_Zero 18:13, 27 February 2018 (UTC)

No, consecutive primes. The first one: 1st and 2nd primes, then 2nd and 3rd, then 3rd and 4th, etc. The second one: 1st and 2nd primes, then 3rd and 4th, then 5th and 6th, etc. Bubba73 ^{You talkin' to me?} 18:24, 27 February 2018 (UTC)