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

= 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)
 * 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. --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)