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This is actually a question which was deleted by a previous user. I worked very hard on this question and became quite engrossed. The question was simply: Is there a closed form for the series $$\sum_{k=1}^\infty \frac{\ln(4k-3)}{(4k-3)}-\frac{\ln(4k-1)}{(4k-1)}?$$

I am submitting my work below as an answer and I'd like Math.SE to help me out here. There is likely something I've missed, and I would like some help on figuring out if there is a closed form.

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    $\begingroup$ Maple did not find a closed form for this series, and maple is usually good about finding such closed forms. This doesn't guarantee there isn't one, but indicates such a closed form likely couldn't be found by some typical "summation trick". $\endgroup$
    – coffeemath
    Dec 8, 2012 at 14:25
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    $\begingroup$ Yup. Mathematica doesn't know how to do it either. $\endgroup$
    – Jackson Walters
    Dec 8, 2012 at 16:17
  • $\begingroup$ @coffeemath Yes, I checked it with Wolfram Alpha. I don't know, but I think W|A and Mathematica have the same set of techniques. That is why it intimidated me; that is, the notion of finding a closed form via techniques known to me. $\endgroup$
    – 000
    Dec 8, 2012 at 20:14

4 Answers 4

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We can rewrite the given sum as $$ \begin{align*} - \sum_{n=0}^{\infty} (-1)^{n+1} \frac{\log(2n+1)}{2n+1} &= - \left. \sum_{n=0}^{\infty} (-1)^{n+1} \frac{\log(2n+1)}{(2n+1)^s} \right|_{s=1} \\ &= - \left. \frac{d}{ds} \right|_{s=1} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^s} \\ &= - \beta'(1) \\ &= - \frac{\pi}{4} \Bigl\{\gamma + 2 \log 2 + 3 \log \pi - 4 \log\left[\Gamma(1/4)\right]\Bigr\}, \end{align*} $$ where $\beta$ is the Dirichlet beta function. The last equality was found at MathWorld, which cites an OEIS entry (among other things), which in turn cites the MathWorld article.

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    $\begingroup$ This matches my computed value (now that I have used Sum instead of NSum) to 50 places. (+1) $\endgroup$
    – robjohn ♦
    Dec 8, 2012 at 17:16
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    $\begingroup$ This is not in the Inverse Symbolic Calculator. One of us should suggest it be added ($\beta'(1)$ is pretty simple not to be included). $\endgroup$
    – robjohn ♦
    Dec 8, 2012 at 17:24
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    $\begingroup$ @Elias, strictly speaking, there is no strict definition of "closed form". $\endgroup$
    – Antonio Vargas
    Dec 8, 2012 at 19:36
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    $\begingroup$ Most people consider something to be in a "closed form" if it is written in terms of "well known" functions (another phrase which is not strictly defined). I would consider the logarithm, pi, Euler's gamma constant, and the gamma function to be "well known". $\endgroup$
    – Antonio Vargas
    Dec 8, 2012 at 19:40
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    $\begingroup$ @AntonioVargas This is an acceptable closed form. I hope it was understood that my question was: "If we can't find it in terms of elementary functions, can we find it in terms of special functions?" While I would love one in terms of elementary functions, it doesn't appear that it exists. In fact, the fact that we have an expression for it in terms of special functions makes me quite skeptical that it can be expressed in terms of elementary functions unless there are some weird things going on with $\log, \Gamma, \pi, \log(\pi),$ and $\gamma$. This is a great answer, Antonio. :-) $\endgroup$
    – 000
    Dec 8, 2012 at 20:12
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While not conclusive proof that there doesn't exist a closed form for this sum, this indicates that there is no simple closed form. The value computed might also act as a hint to the (in)validaty of a closed form.

This series converges very slowly. Using the Euler-Maclaurin Sum Formula, we get the asymptotic expansion:

$$ \begin{array}{l} \sum_{k=1}^n\frac{\log(4k-a)}{4k-a}\\ \sim\small\left(\frac1{3(4n{-}a)^2}-\frac{44}{45(4n{-}a)^4}+\frac{8768}{945(4n{-}a)^6}-\frac{30976}{175(4n{-}a)^8}+\frac{58400768}{10395(4n{-}a)^{10}}\right)\\ \small+\left(\frac{1}{2(4n{-}a)}-\frac{1}{3 (4n{-}a)^2}+\frac{8}{15 (4n{-}a)^4}-\frac{256}{63 (4n{-}a)^6}+\frac{1024}{15 (4n{-}a)^8}-\frac{65536}{33 (4n{-}a)^{10}}\right) \log (4n{-}a)\\ \small+\frac{1}{8} \log ^2(4n{-}a)+C_a \end{array} $$ Note that all the terms vanish as $n\to\infty$ except the last two. Furthermore, $$ \begin{align} 0 &\le\lim_{n\to\infty}\left(\log^2(4n-1)-\log^2(4n-3)\right)\\[9pt] &=\lim_{n\to\infty}\left(\log(4n-1)+\log(4n-3)\right)\left(\log(4n-1)-\log(4n-3)\right)\\ &\le\lim_{n\to\infty}2\log(4n-1)\log\left(1+\frac2{4n-3}\right)\\ &\le\lim_{n\to\infty}\frac{4\log(4n-1)}{4n-3}\\[6pt] &=0 \end{align} $$ Therefore, $$ \sum_{k=1}^\infty\frac{\log(4k-3)}{4k-3}-\frac{\log(4k-1)}{4k-1}=C_3-C_1 $$ Plugging in $n=10000$ into the asymptotic expansion for $a=1$ and $a=3$, we get to $50$ places $$ \begin{align} C_1&=+0.03827947109192929052721132046406192021405884328730\\ C_3&=-0.15462184570498313883597844356397086503103802433270\\ C_3-C_1&=-0.19290131679691242936318976402803278524509686762001 \end{align} $$ Using the Inverse Symbolic Calculator, there doesn't seem to be a simple expression for this number.

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  • $\begingroup$ The last digit is off by 1 due to rounding in the subtraction (I carried out the computation to 60 places). $\endgroup$
    – robjohn ♦
    Dec 8, 2012 at 17:20
  • $\begingroup$ I think your answer is very valuable even though it does not provide a closed form. +1. $\endgroup$
    – 000
    Dec 8, 2012 at 20:15
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$$\sum_{k=1}^\infty \frac{\ln(4k-3)}{(4k-3)}-\frac{\ln(4k-1)}{(4k-1)}=\sum_{k}\left(\frac{\ln(4k-3)}{4k-3}-\frac{\ln(4k-1)}{4k-1} \right)[k \ge 1].$$

Let $k=j+1$. As a result, $$4k-3=4j+4-3=4j+1, 4k-1=4j+4-1=4j+3, \text{ and } k\ge 1 \implies j \ge 0.$$

$$\sum_{k}\left(\frac{\ln(4k-3)}{4k-3}-\frac{\ln(4k-1)}{4k-1} \right)[k \ge 1]=\sum_j \left( \frac{\ln(4j+1)}{4j+1}-\frac{\ln(4j+3)}{4j+3}\right)[j \ge 0].$$

There is no 'nice' cancelling that occurs between these two summands. This is because the set $A=\{4j+1: j\ge 0\}$ and $B=\{4j+3: j \ge 0\}$ have no elements of intersection; that is, $A\cap B=\emptyset.$ While it isn't entirely relevant, this can be formally proven by noting $A$ and $B$ are equivalence classes $[1]=\{b \in \mathbb{Z}:b\equiv 1 \pmod 4\}$ and $[3]=\{b \in \mathbb{Z}: b\equiv 3\pmod 4\}$, respectively. $[1]$ and $[3]$ partition $\mathbb{Z}$ into disjoint sets, hence $A\cap B=\emptyset$. That begs the question: Where to from here?

To be honest, I don't know. I cannot conclusively prove that there is not a closed form solution for this series, but I certainly cannot find it. It does converge, and the integral test provides a nice upperbound for it.

I hope someone else can be of assistance.

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  • $\begingroup$ I think there is tricky way to do that, if we accept we can find a close form for it. $\endgroup$
    – Mikasa
    Dec 8, 2012 at 11:42
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Hint. For all sequence $\{b_k\}_{k\in\mathbb{N}}$ we have \begin{equation} \begin{split} \sum_{k=1}^\infty \frac{\ln(4k-3)}{(4k-3)}-\frac{\ln(4k-1)}{(4k-1)} & = \sum_{k=1}^\infty \left[\frac{\ln(4k-3)}{(4k-3)}+ b_k\right] - \left[ \frac{\ln(4k-1)}{(4k-1)} + b_k\right] \\ \end{split} \end{equation} Supose that $\sum_{k=1}^\infty b_k$ converge and
$$ a_{k+1}= \left[ \frac{\ln(4k-3)}{(4k-3)}+b_k\right] \quad a_{k}= \left[\frac{\ln(4k-1)}{(4k-1)}+ b_k\right] $$ Use the telescopic propert of series: $\sum_{k=1}^{\infty} (a_{k+1}-a_k)= (\lim_{k\to\infty}a_{k+1})- a_1$.

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    $\begingroup$ Summing the first million terms yields -0.192903. $\frac{1}{2}\ln(2)-\frac{1}{3}\ln(3)=-0.0196305$. Using your definitions of $a_{k}$ and $a_{k+1}$, the original series is $\displaystyle \sum_{k=1}^{\infty} a_{k}+a_{k+1}$, not $\displaystyle \sum_{k=1}^{\infty} a_{k+1}-a_{k}.$ $\endgroup$
    – Jackson Walters
    Dec 8, 2012 at 16:12
  • $\begingroup$ @Jackson Walters That was a mistake on my part. But I will update my answer. I think the reasoning is right on. $\endgroup$
    – Elias Costa
    Dec 8, 2012 at 18:20
  • $\begingroup$ This was a good attempt at the problem, but I pointed out that it would appear there is no telescoping. I don't know what your idea was, precisely, but it's good that you had the gall to give it a shot. $\endgroup$
    – 000
    Dec 8, 2012 at 20:18

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