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I'm looking for a simplification of $$ \sum _{k=r+1}^{2 r} \frac{2 k+2 r+1}{2 k^2-k (2 r+1)+2 r (r+1)}\:. $$ Mathematica gives a somewhat tautological result in terms of the digamma function $\psi$: $$ \begin{align*} \frac{1}{2} \left((6 r+R+3) \left(\psi\left(\frac{1}{4} (6 r-R+3)\right)-\psi\left(\frac{1}{4} (2 r-R+3)\right)\right)+(6 r-R+3) \left(\psi\left(\frac{1}{4} (2 r+R+3)\right)-\psi\left(\frac{1}{4} (6 r+R+3)\right)\right)\right)\:, \end{align*} $$ where $R=\sqrt{1-12r-12r^2}$. However, on a related sum, Mathematica output a similar answer where the true answer could be written much simpler, so I'm not sure this is the best possible.

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Consider first $$\frac{2k+a}{2(k-b)(k-c)}=\frac 1{2(b-c)}\left(\frac{a+2b}{k-b}- \frac{a+2c}{k-c}\right)$$ where $$a=2r+1 \qquad b=\frac{1}{4} \left(2r+1\sqrt{-12 r^2-12 r+1}\right)\qquad c=\frac{1}{4} \left(2r+1-\sqrt{-12 r^2-12 r+1}\right)$$ and now $$\sum _{k=r+1}^{2 r} \frac1{k-b}=H_{2 r-b}-H_{r-b}\qquad \sum _{k=r+1}^{2 r} \frac1{k-c}=H_{2 r-c}-H_{r-c}$$ where appear harmonic numbers.

So $$\sum _{k=r+1}^{2 r}\frac{2k+a}{2(k-b)(k-c)}=\frac{(a+2 b) \left(H_{2 r-b}-H_{r-b}\right)-(a+2 c) \left(H_{2 r-c}-H_{r-c}\right)}{2 (b-c)}$$

Do you prefer this one ?

Edit

Just for your curiosity, using the asymptotics of harmonic numbers, when $r$ is large, the summation is $$\frac{1}{6} \left(\sqrt{3} \pi +3\log (3)\right)-\frac{1}{2 r}+\frac{17+2 \sqrt{3} \pi }{72 r^2}-\frac{11+2 \sqrt{3} \pi }{72 r^3}+\frac{133+60 \sqrt{3} \pi }{2160 r^4}+O\left(\frac{1}{r^5}\right)$$ which seems to be quite accurate as shown below $$\left( \begin{array}{ccc} r & \text{exact} & \text{approximation} \\ 5 & 1.369565848 & 1.369605205 \\ 10 & 1.409794490 & 1.409795782 \\ 15 & 1.424507630 & 1.424507803 \\ 20 & 1.432137276 & 1.432137318 \\ 25 & 1.436806524 & 1.436806537 \\ 30 & 1.439958450 & 1.439958456 \\ 35 & 1.442229294 & 1.442229297 \\ 40 & 1.443943198 & 1.443943199 \\ 45 & 1.445282671 & 1.445282672 \\ 50 & 1.446358333 & 1.446358333 \end{array} \right)$$

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  • $\begingroup$ The issue would be that $b,c$ are complex ($r\ge1$), so harmonic numbers aren't quite the correct notion (as far as terms like $H_{r-b}$ are concerned). $\endgroup$ – Ross B. Nov 1 '17 at 5:49
  • $\begingroup$ @RossB. What about the generalized harmonic numbers ? $\endgroup$ – Claude Leibovici Nov 1 '17 at 5:55
  • $\begingroup$ I think that would be fine, but I think what you wrote is essentially the digamma version that Mathematica returned. $\endgroup$ – Ross B. Nov 1 '17 at 5:58
  • $\begingroup$ @RossB.. I agree but I feel it is nicer and easy to make more general for $a(r)$, $b(r)$, $c(r)$. $\endgroup$ – Claude Leibovici Nov 1 '17 at 5:59

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