EDIT, Will Jagy, December 8, 2010: to anyone considering working on this, please first see http://mathoverflow.tqft.net/discussion/817/could-a-few-moderators-please-remove-one-of-my-questions/#Item_9 which gives the story behind this peculiar sum. Note that the OP is no longer interested in the results, as they arose from one kind of error and cannot be applied because of a different sort of misunderstanding. The double sum version below was provided recently by Harald Hanche-Olsen.

ORIGINAL. I'm curious one of you is able to find the exact evaluation of the following series:

$$\begin{aligned} S &= 1/(2\times3) +1/(5\times6) + 1/(7\times8) + 1/(10\times11) + \cdots \\\\&= \sum_{n=1}^\infty\sum_{k=1}^{n}\frac1{(n^2+2k-1)(n^2+2k)} \end{aligned}$$

I'm not exactly sure on how to state the 'general term' of the series. Perhaps I can illustrate it with an example:

$ 1/(1\times2) + 1/(3\times4) + 1/(5\times6) + 1/(7\times8) + \ldots + 1/((2n - 1) \times 2n) + \ldots = \log(2)$.

Now, to answer Nate Eldredge: let $a_0=2$ and $a_{k+1}=a_{k} + 1 $ unless $ a_{k} + 1$ is a square, in which case let $a_{k + 1} = a_{k} + 2$. Now, multiply $a_{k}$ with $a_{k+1}$. That's a term. Let me show the first few terms:

$ S = 1/(2\times3)$ [now skip 4] $ + 1/(5\times6) + 1/(7\times8)$ [now skip 9] $ + 1/(10\times11) + 1/(12\times13) + 1/(14\times15)$ [now skip 16] $ + 1/(17\times18) + \ldots $

So all the squares (1,4,9,16,25, etc) are 'skipped' in the terms.

I hope this clarifies it a bit...

Thanks a lot in advance,

Max Muller

PS: If someone has any ideas as to how the general term of this series can be written in a more concise manner, please let me know! For the Meta-users, see also the relevant discussion on this question.

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    $\begingroup$ Is there a mistake in your series? It's not quite clear how the general term relates to your first few terms. $\endgroup$ – gowers May 26 '10 at 16:13
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    $\begingroup$ So to clarify, let $a_0 = 2$, $a_{k+1} = a_k + 1$ if $a_k + 1$ is not a perfect square, $a_{k+1} = a_k + 2$ otherwise. You are asking about $S = \sum_{k=0}^\infty 1/(a_{2k} a_{2k+1})$. Have I got it right? $\endgroup$ – Nate Eldredge May 26 '10 at 16:20
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    $\begingroup$ Dealing with those omitted terms is not standard calculus, although there are standard techniques to do it. $\endgroup$ – David E Speyer May 26 '10 at 16:53
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    $\begingroup$ \times is the LaTeX code for the multiplication cross; $1 \times 2$ looks much nicer than $1x2$. If you want a multiplication dot, it's \cdot. $\endgroup$ – JBL May 27 '10 at 0:13
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    $\begingroup$ Mathematica reports the sum as being close to 0.2666107465545, which doesn't seem to be in the inverse calculator. No time for a smart lookup... $\endgroup$ – Kevin O'Bryant Dec 5 '10 at 18:56

I thought I'd add in my favorite way of computing sums of the form $$\sum_{n=0}^{\infty} \frac{p(n)}{q(n)}$$ for $p$ and $q$ polynomials.

First, expand in partial fractions: $$\frac{p(n)}{q(n)} = \sum \frac{a_i}{n+s_i} + \mbox{terms for repeated roots}.$$

Since the sum is to converge, we must have $p(n)/q(n) =O(n^{-2})$, so $\sum a_i=0$. So rewrite this as $$\frac{p(n)}{q(n)} = \sum a_i \left( \frac{1}{n+s_i} - \frac{1}{n} \right) + \mbox{terms for repeated roots}.$$

We are now reduced to evaluating sums of the form $$\sum \left( \frac{1}{n} - \frac{1}{n+s} \right)$$ and $$\sum \frac{1}{(n+s)^k}$$

We have the identity $$\sum \left( \frac{1}{n} - \frac{1}{n+s}\right) = \frac{\Gamma'(s)}{\Gamma(s)} + \gamma + \frac{1}{s}$$ where $\Gamma$ is the Gamma function and $\gamma$ is the Euler–Mascheroni constant. See, for example, this website. Taking repeated derivatives of this gives a formula for $\sum 1/(n+s)^k$.

So, if you accept derivatives of the $\Gamma$ function at algebraic numbers, such sums can always be evaluated. In general, I don't know a better method. However, in many cases, one can use familiar $\Gamma$ function identities to do better. Here are three tricks, all of which come up in our setting:

  1. If we ever have to deal with $\sum \left( \frac{1}{n+s} - \frac{1}{n+k+s} \right)$, then the sum telescopes. We can see this on the $\Gamma$ function side: $\Gamma(s+k) = (s+k-1)\cdots (s+1) (s) \Gamma(s)$ and taking logarithimic derivatives gives a relation between $\Gamma(s+k)/\Gamma'(s+k)$ and $\Gamma'(s)/\Gamma(s)$.

  2. If we ever have to deal with $\sum \left( \frac{1}{n-s} - \frac{1}{n+s} \right)$, recall that $$\Gamma(s) \Gamma(-s) = \frac{\pi}{s \sin (\pi s)}.$$ Taking logarithimic derivatives of this will give you a formula for $\Gamma'(s)/\Gamma(s) - \Gamma'(-s)/\Gamma(-s)$. Repeated derivatives will deal with $\sum \left( \frac{1}{(n-s)^k} - \frac{1}{(n+s)^k} \right)$. It is important to note that all of these formulas work for $s$ complex; for example, in the current example we need to deal with $s=i/2$.

  3. Remember that $\zeta(2k)$ is already known, and is easy to look up that than the Taylor series of $\Gamma$ around $0$.

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  • $\begingroup$ @ David: although it doesn't adress the question, I very much appreciate the fact that you have given me this information. Thanks! $\endgroup$ – Max Muller Dec 9 '10 at 19:39

NB: It seems that after this answer was written, the asker "made precise" what he wanted, and I think this has been rendered more or less irrelevant...

NB2: Maybe the people voting down this can explain their votes?

Assuming you want to sum the terms of $\displaystyle\sum_{n\geq1}\frac1{n(n+1)}$ which are not of the form $\displaystyle\frac{1}{n^2(n^2+1)}$ or $\displaystyle\frac{1}{(n^2-1)n^2}$, you can do it by computing the first sum (by the telescoping trick) and then substracting the sum of the terms you want to exclude. To compute the sums you need to substract, say, $\displaystyle\sum_{n\geq2}\frac{1}{(n^2-1)n^2}$, you can compute it as the sum of residues of $\frac{\cot z}{z^4-z^2}$ at $2$, $3$, $4$, etc, using a bit of complex analysis (as in the 8th solution of the Basel problem presented by R. Chapman here)

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  • $\begingroup$ But the point seems to be to "restart" after each square, adding number of terms and stopping? So that the general term is misleading as given. $\endgroup$ – Charles Matthews May 26 '10 at 16:53
  • $\begingroup$ No, I don't think I want to omit the terms of the form $1/(n^2(n^2+1))$ or 1/(n^2(n^2-1)) . I don't think this is some classical problem of a calculus textbook. Let me edit my question a bit, perhaps I can clarify it. $\endgroup$ – Max Muller May 26 '10 at 16:53
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    $\begingroup$ We have $1/(n^2(n^2-1))=-1/n^2 + 1/2(n-1) - 1/2(n+1)$, so this sum is $latex -\zeta(2)$ plus a telescoping term. I think that the other omitted term genuinely requires the residue trick, though. $\endgroup$ – David E Speyer May 26 '10 at 16:56
  • $\begingroup$ Indeed, Mathematica sums that one to $\frac{7}{4}-\frac{\pi ^2}{6}$, which screams zeta, while the other one to $\frac{1}{6} \pi (\pi -3 \coth (\pi ))$, which screams residues :) $\endgroup$ – Mariano Suárez-Álvarez May 26 '10 at 17:01
  • $\begingroup$ Please reformulate with a -1 to the power of floor function of square root of n, applied to alternating reciprocals, and omitting the squares. Then I think we might understand. "Classical" does not mean calculus textbook, necessarily, though. $\endgroup$ – Charles Matthews May 26 '10 at 17:05

As already mentioned by Mariano, the sum in question is a simple logarithmic sum minus $$ \sum_{n>1}\biggl(\frac1{n^2(n^2-1)}+\frac1{n^2(n^2+1)}\biggr) =2\sum_{n>1}\frac1{n^4-1} =\frac74-\frac\pi2\coth\pi. $$ The closed form evaluation is mentioned in Michel Waldschmidt's "Open Diophantine Problems" (Moscow Math. Journal 4:1 (2004), 245-305, 312) together with "which is a transcendental number since $\pi$ and $e^\pi$ are algebraically independent over $\mathbb Q$ (Yu.V. Nesterenko)."

EDIT. I was confused by too many answers and edits. On using $$ \frac1{n(n+1)}=\frac1n-\frac1{n+1}, $$ one can write the wanted sum as $$ \sum_{n=1}^\infty\frac{(-1)^{n+[\sqrt n]-1}}n+\sum_{k=1}^\infty\frac1{k^2} =\sum_{n=1}^\infty\frac{(-1)^{n+[\sqrt n]-1}}n+\frac{\pi^2}6. $$ I can hardly imagine that the remaining single sum has a closed form... But this leaves a nice problem in 1st year analysis: Show that the series $$ \sum_{n=1}^\infty\frac{(-1)^{n+[\sqrt n]}}n $$ converges.

  • $\begingroup$ I think the answer has been found, finally! Thanks, Wadim Zudilin. $\endgroup$ – Max Muller Dec 6 '10 at 14:16
  • $\begingroup$ Hi Wadim,I don't think you've read the question correctly (but it's hard to tell because I don't know precisely what you mean by "a simple logarithmic sum"). Each non-square positive integer $i$ appears here as a factor in the denominator of exactly one term of the sum. $\endgroup$ – JBL Dec 6 '10 at 14:23
  • $\begingroup$ (You'll see that this sum was already worked out by Mariano in the 4th comment on Mariano's answer.) $\endgroup$ – JBL Dec 6 '10 at 14:24
  • $\begingroup$ @ Wadim: I'm afraid JBL is right, I'm sorry I accepted and then 'unaccepted' your answer. $\endgroup$ – Max Muller Dec 6 '10 at 14:52
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    $\begingroup$ @Wadim: That is a nice problem. +1 $\endgroup$ – Andrés E. Caicedo Dec 9 '10 at 6:24

rmk 1. After some manipulation of the series, if

$f(x):=1 - \frac{1-x}{1+x}\sum_{k\in\mathbb Z} x^{k^2}$


$S:=\frac{1}{2}\int_0^1\frac{f(x)}{x} dx.$

So this confirms Charles Matthews' hint:

$\theta:=\sum_{k\in\mathbb Z} x^{k^2}$ Is the theta series, indeed, linked to the Jacobi theta function. (But I don't know how to proceed then.)

rmk 2. Maybe writing Max Muller's series as a sum over double indices (k,n)

$S:=\sum_{1\le k\le n} \frac{1}{(n^2+2k-1)(n^2+2k)}$

one can find a smart partition of the set of indices that allows an iterate summation. (very few chances of success). For instance changing the order of summation, i.e.

$S:=\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{1}{(n^2+2k-1)(n^2+2k)}$

The inner sum can be treated via partial fraction decomposition and summed as illustrated by David Speyer; I found (with $\Psi:=\Gamma'/\Gamma$)

$\sum_{k=1}^\infty\ \left( \mathrm{Im}\frac{\Psi(k+i\sqrt{2k-1})}{\sqrt{2k-1}}-\mathrm{Im}\frac{\Psi(k+i\sqrt{2k})}{\sqrt{2k}}\right)=0.2666107..$

  • $\begingroup$ Thanks, Pietro. But isn't a 'double sum' a sum of a sum? So shouldn't there be two sigmas in rmk 2? $\endgroup$ – Max Muller May 27 '10 at 14:40
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    $\begingroup$ To be precise, it's a single sum over a set of double indices (k,n) :-) , that you can sum in any order and with any partition you like, since all terms are positive. I added a few lines that may be of some interest to make a quicker numeric evaluation. $\endgroup$ – Pietro Majer May 27 '10 at 15:00
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    $\begingroup$ @Pietro : I think your integral representation of the sum gives a very fast way to evaluate $S$ numerically, using the functional equation of $\theta$. This seems to give a double sum but this time with an exponentially decreasing factor. $\endgroup$ – François Brunault Dec 6 '10 at 16:35
  • $\begingroup$ Interesting! will you please add some detail? $\endgroup$ – Pietro Majer Dec 6 '10 at 22:10

As you are interested in $ \zeta(3) $ you might prefer this variant of your construction. This is also a small part of what Pietro Majer would have done in the direction indicated by Charles Matthews.

Define $ f(x)$ for $ | x | \leq 1 $ by

$$ f(x) = \frac{x^4}{2 \cdot 3 \cdot 4} + \frac{x^7}{5 \cdot 6 \cdot 7} + \frac{x^{11}}{9 \cdot 10 \cdot 11} + \frac{x^{14}}{12 \cdot 13 \cdot 14} +\cdots + \frac{x^{30}}{28 \cdot 29 \cdot 30} + \frac{x^{33}}{31 \cdot 32 \cdot 33} + \cdots $$

Then I took the third derivative, power series has all coefficients 1 and simplifies as

$$ f'''(x) = (x + x^4 ) + ( x^8 + x^{11} + \cdots + x^{23}) + ( x^{27} + x^{30} + \cdots) + \cdots $$

$$ f'''(x) = \sum_{n=1}^\infty \; \; x^{n^3} \left( \frac{1 - x^{3 n^2 + 3 n}}{1 - x^3} \right) $$ or

$$ \left( \frac{1 }{1 - x^3} \right) \cdot \left( \sum_{n=1}^\infty \; \; x^{n^3} - x^{n^3 + 3 n^2 + 3 n} \right) $$

$$ \left( \frac{1 }{1 - x^3} \right) \cdot \left( \sum_{n=1}^\infty \; \; x^{n^3} - \left( \frac{1}{x} \right) \sum_{n=1}^\infty \; \; x^{(n + 1)^3} \right) $$

$$ \left( \frac{1 }{1 - x^3} \right) \cdot \left( \sum_{n=1}^\infty \; \; x^{n^3} - \left( \frac{1}{x} \right) \sum_{m=2}^\infty \; \; x^{m^3} \right) $$

$$ \left( \frac{1 }{1 - x^3} \right) \cdot \left( 1 + \sum_{n=1}^\infty \; \; x^{n^3} - \left( \frac{1}{x} \right) \sum_{m=1}^\infty \; \; x^{m^3} \right) $$

$$ \left( \frac{1 }{1 - x^3} \right) \cdot \left( 1 - \left( \frac{1-x}{x} \right) \sum_{m=1}^\infty \; \; x^{m^3} \right) $$

$$ f'''(x) = \left( \frac{1 }{1 - x^3} \right) - \left( \frac{1}{x ( 1 + x + x^2)} \right) \sum_{m=1}^\infty \; \; x^{m^3} $$

And so on more or less forever, with no explicit value for $f(1)$ likely.

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  • $\begingroup$ I don't know how this came out community wiki. Must have clicked something i did not notice. $\endgroup$ – Will Jagy May 29 '10 at 3:02
  • $\begingroup$ The question was converted to community wiki due to the large number of edits, and so all new answers are automatically CW as well. $\endgroup$ – j.c. May 29 '10 at 17:31
  • $\begingroup$ Thanks, I did not think of any cause of that type. $\endgroup$ – Will Jagy May 30 '10 at 22:16

Anyway, I think the point is that with a(k) a sequence of the type implicated here, namely blocks of +1 and -1 alternating, jumping and flipping over at k any square, the generating function of the a(k) is composed of blocks of geometric series. If the variable is t, each finite geometric series can be summed with denominator 1 + t. At this stage I think Euler could handle the question? Rational function times a theta-series, I believe. This needs to be integrated formally, and the limit as t -> 1 from below taken. This would be the "classical" approach.

I apologise if this is nonsense.

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  • $\begingroup$ I'm not sure if I have understood your 'classical approach... Could you describe what you're doing in 'mathematical language?'. And why do you think Euler could handle this question? Has he evaluated this series already? Please ellaborate on your methods, anyway! $\endgroup$ – Max Muller May 26 '10 at 20:10
  • $\begingroup$ I have to say that I'm not going to type more mathematical notation here, given your reaction above. It is a bit ironic that you ask for an explanation in mathematical language. It is not clear to me whether this is a challenge problem to which you already know the answer, but I'm not highly motivated yet (whether it is or not). You have ignored the request I made to formulate the series in sigma notation: "..." can waste a lot of time, as the discussion shows. $\endgroup$ – Charles Matthews May 26 '10 at 20:59
  • $\begingroup$ I'm sorry I didn't formulate the series in sigma notation, I'm not very familiar with LaTeX and I more or less forgot to listen to your request. Of course I know your speaking in 'matematical language', I was thinking more of symbols and an exact proof. I was just interested in your approach and I tried to encourage you to expand your answer, as mister Alvarez probably gave a nice and correct answer, but not a very 'clean proof'. I think it's nice to have more than one solution to a problem, but if you don't want to type any more mathematical notation anymore, I understand. $\endgroup$ – Max Muller May 26 '10 at 21:19
  • $\begingroup$ You realise that Alvarez adds 1/5x6, 1/6x7, 1/7x8, 1/10x11, 1/11x12 etc.? $\endgroup$ – Charles Matthews May 26 '10 at 21:59
  • $\begingroup$ No.. I didn't realize that. I thought he added 1/5x6, 1/7x8, 1/10x11 etc., which is what I would've liked. I had some doubts, but Michael Greenblatt convinced me. $\endgroup$ – Max Muller May 26 '10 at 22:15

Incorrect answer deleted.

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    $\begingroup$ But what is left is not Max Muller's series either :-( the terms 1/2x3 or 1/10x11 are still lacking, while there are extra terms as 1/13x14 not included in Max's series. $\endgroup$ – Pietro Majer May 27 '10 at 11:57
  • $\begingroup$ Grr, you're right. He keeps doing things which seem bizarre. OK, I'll rethink this. $\endgroup$ – David E Speyer May 27 '10 at 12:51
  • $\begingroup$ I guess you're right Will. I'm young and inexperienced but I'm also willing to work and think hard to understand things. But did you say this because David Speyer said 'he keeps doing things which seem bizarre'? Because I'm not sure David was referring to me when he said 'he'.. or are you, David? $\endgroup$ – Max Muller May 27 '10 at 19:47
  • $\begingroup$ I was. Of course, I have no ideal what the motivation for this sum is. But the way you are dropping out terms means that you are switching between sums of the form $\sum 1/((2k)(2k+1))$ and $\sum 1/((2k-1)(2k))$. I think I kept misreading the definition because that seemed like a very difficult and unmotivated thing to do; if I were just making up a sum for the fun of it, I wouldn't do that. $\endgroup$ – David E Speyer May 27 '10 at 19:50
  • $\begingroup$ I hope this doesn't come off as hostile. It wasn't meant to be, and I will be interested to hear if anyone finds a closed form for your sum $\endgroup$ – David E Speyer May 27 '10 at 19:50

S=1/(2×3)+1/(5×6)+1/(7×8)+1/(10×11)+...=1/20(20-2(pi)Sqrt(5-2Sqrt(5)) -2Sqrt(5)ArcCoth(Sqrt(5))-5Log(5))=0.2617882198...(approx.)

  • $\begingroup$ The name of a real number with no explanation where it came from is of close to 0 value. $\endgroup$ – JBL Dec 5 '10 at 18:19
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    $\begingroup$ Can you give a proof? $\endgroup$ – Nate Eldredge Dec 5 '10 at 21:40
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    $\begingroup$ The sum in the question has value more than 0.2666 , so this answer is wrong. I get 0.2666 by adding terms up to 500^2. That is, up to $1/(249998\times249999)$. And the remaining terms are positive, so the complete sum is more than this. $\endgroup$ – Gerald Edgar Dec 6 '10 at 0:04
  • $\begingroup$ Here is a way to write this sum, each term of the original is a + - pair in this. $$ \frac{1}{2}\;\sum_{k = 1}^{\infty} \frac{(-1)^{(k + \mathrm{ceil} (\sqrt{k}))} - (-1)^{(k + \mathrm{floor} (\sqrt{k}))}}{k} $$ $\endgroup$ – Gerald Edgar Dec 6 '10 at 0:07

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