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I need a closed form for the sum $S_m$ denoted as $$S_m = \sum_{k=1}^{m} \frac{(-1)^{k+1} (m-k) \ c_{m-k}}{(2k+1)!}\tag{1}$$ where $c_0=1$ and $$c_m= \frac{(-1)^m}{(2m+1)!}+\frac{1}{m}\sum_{k=1}^m \frac{(-1)^{k+1}(m-k)\ c_{m-k}}{(2k+1)!} \ \ \ \ \ \ \ \ \ \forall\ \ \ m\in\mathbb{N}\tag{2}$$

I tried the following $$c_{m-k}= \frac{(-1)^{m-k}}{(2m-2k+1)!}+\frac{1}{m-k}\sum_{k=1}^{m-k} \frac{(-1)^{k+1}(m-2k)\ c_{m-2k}}{(2k+1)!} \ \ \ \ \ \ \ \ \ \forall\ \ \ m\in\mathbb{N}\tag{3}$$ So by $(1)$ and $(3)$ $$S_m=\sum_{k=1}^{m} \frac{ (-1)^{m+1}(m-k)}{(2k+1)!(2m-2k+1)!}+\sum_{k=1}^{m}\sum_{k=1}^{m-k} \frac{ (-1)^{2k+2}(m-2k)\ c_{m-2k}}{((2k+1)!)^2} \tag{4}$$ Again by $(3)$ $$c_{m-2k}= \frac{(-1)^{m-2k}}{(2m-4k+1)!}+\frac{1}{m-2k}\sum_{k=1}^{m-2k} \frac{(-1)^{2k+1}(m-4k)\ c_{m-4k}}{(4k+1)!}\tag{5}$$ By $(4)$ and $(5)$ we have $$S_m=\sum_{k=1}^{m} \frac{(-1)^{m+1}(m-k)}{(2k+1)!(2m-2k+1)!}+\sum_{k=1}^{m}\sum_{k=1}^{m-k}\frac{(-1)^{m+2}(m-2k)}{((2k+1)!)^2(2m-4k+1)!}+ \sum_{k=1}^{m}\sum_{k=1}^{m-k}\sum_{k=1}^{m-2k}\frac{(-1)^{4k+3}(m-4k)\ c_{m-4k}}{((2k+1)!)^2(4k+1)!}$$ Write in the above equation $$S_m=S^1_m+S^2_m+S^3_m $$ $$S^1_m=\sum_{k=1}^{m} \frac{(-1)^{m+1}(m-k)}{(2k+1)!(2m-2k+1)!}$$ $$S^1_m=(-1)^{m+1} m\sum_{k=1}^{m} \frac{1}{\Gamma(2k+2)\Gamma(2m-2k+2)}+(-1)^{m} \sum_{k=1}^{m} \frac{k}{\Gamma(2k+2)\Gamma(2m-2k+2)} $$ Now we have by Wolfram alpha (see : https://www.wolframalpha.com/input?i=sum+1%2F%28Gamma%282k%2B2%29+Gamma%282m-2k%2B2%29%29%2C+k%3D1+to+m) $$(-1)^{m+1} m\sum_{k=1}^{m} \frac{1}{\Gamma(2k+2)\Gamma(2m-2k+2)}= \frac{2 (-1)^{m} m (m-4^m+1)}{\Gamma(2m+3)}$$ For the next piece we have (see:https://www.wolframalpha.com/input?i=sum+k%2F%28Gamma%282k%2B2%29+Gamma%282m-2k%2B2%29%29%2C+k%3D1+to+m ) $$(-1)^{m} \sum_{k=1}^{m} \frac{k}{\Gamma(2k+2)\Gamma(2m-2k+2)}=\frac{(-1)^m 4^m m}{\Gamma(2m+3)}$$ So we get $$S^1_m=\frac{(-1)^m m(2(m+1)-4^m)}{\Gamma(2m+3)} $$ I am struggling to find the closed form for the sum in question. Any help would be highly appreciated. Thank you.

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  • $\begingroup$ Do you not need to have a different summation indices for (1) and (2)? You certainly can't mean to have $k$ for both. $\endgroup$
    – JimB
    Mar 7, 2023 at 5:58
  • $\begingroup$ @JimB Thank you for your comment. Sorry I can't understand that But why we have to change the $k$ in one of $(1)$ or $(2)$? $\endgroup$
    – Max
    Mar 7, 2023 at 7:12
  • $\begingroup$ @AaronHendrickson Thank you for your comment. Can we have a closed form of this sum? $\endgroup$
    – Max
    Mar 7, 2023 at 13:52
  • $\begingroup$ My comment is more important for the double and triple sums. $\endgroup$
    – JimB
    Mar 7, 2023 at 17:51

1 Answer 1

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A "closed form" is in the eye of the beholder.

One can define $c_m$ in a non-recursive fashion:

$$c_m=\frac{2^{2 m-1} (-1)^m B_{2 m}}{m (2 m)!}$$

where $B_n$ is the $n$-th Bernoulli number. This is found by looking at the sequences of the numerators and denominators of values of $c_m$ (calculated by your original formulas) at oeis.org which found sequences A283301 and A046989.

So a "closed form" for $S_m$ is

$$S_m=\sum _{k=1}^{m-1} \frac{(-1)^{m+1} 2^{2 (m-k)-1} B_{2 (m-k)}}{(2 k+1)! (2 (m-k))!}$$

We sum just to $m-1$ because of the term $(m-k)$ is zero when $k=m$.

Then using Mathematica one can define $S_m$ as

s[m_] := Sum[((-1)^(m + 1)*2^(-1 + 2*(m - k))*BernoulliB[2*(m - k)])/
  ((1 + 2*k)!*(2*(m - k))!), {k, 1, m - 1}]

A table of values for $S_m$ are easily found:

Table[{m, s[m]}, {m, 1, 10}] // TableForm

Table of m and S_m

A plot of $m$ vs $\log{|S_m|}$ suggests a linear fit would be a good approximation (maybe starting from $m=10$).

data = Table[{m, Log[Abs[s[m]]] // N}, {m, 2, 300}];
ListPlot[data[[10 ;;]]]

Plot of m vs log(|S_m|)

The suggestion by @GaryMar (at least I think what that suggestion means) is to use

$$\sum _{m=1}^{\infty } S_m x^{2 m}=\frac{(\sin (x)-x) (1-x \cot (x))}{2 x}$$

as a generating function and therefore $S_n$ can be found by

ss[m_] := Limit[D[(1 - x Cot[x]) (Sin[x] - x)/(2 x), {x, 2 m}]/(2 m)!, x -> 0]
ss[5]
(* -(23/2177280) *)

This works but starts to get very slow for values of $m>5$.

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    $\begingroup$ Note that $$ \sum\limits_{k = 1}^\infty {S_k x^{2k} } = \left( {\sum\limits_{k = 1}^\infty {( - 1)^{k + 1} \frac{{2^{2k - 1} B_{2k} }}{{(2k)!}}x^{2k} } } \right)\left( {\sum\limits_{k = 1}^\infty {\frac{{( - 1)^k }}{{(2k + 1)!}}x^{2k} } } \right) = \frac{{1 - x\cot x}}{2}\frac{{\sin x - x}}{x}. $$ You can find the asymptotics for $S_k$ using singularity analysis. $\endgroup$
    – Gary
    Mar 9, 2023 at 1:17
  • $\begingroup$ Thanks for your answer +1 for your effort. Please show a proof of how did you get the following $$c_m=\frac{2^{2 m-1} (-1)^m B_{2 m}}{m (2 m)!}$$ $\endgroup$
    – Max
    Mar 9, 2023 at 5:53
  • $\begingroup$ I don't have a proof other than finding both the numerator and denominator sequences for $c_m$ in oeis.org. Numerators and denominators of $c_m$ are found in sequences A283301 and A046989, respectively. $\endgroup$
    – JimB
    Mar 9, 2023 at 6:21
  • $\begingroup$ Thanks. I looked at OEIS A283301 AND A046989. How do we get the numerator and denominator sequences of $c_m$ in oeis.org? In oeis we have expansion of $\log (x/\sin x)$ $\endgroup$
    – Max
    Mar 9, 2023 at 6:50
  • $\begingroup$ For those algorithm's there are several pieces of code for various languages to get the expansion of $\log{(x/\sin{x})}$. $\endgroup$
    – JimB
    Mar 9, 2023 at 6:53

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