# Backslide of Limit

Backslide introduced in 9.0, and persisting through 12.0.

A friend of mine showed me this example:

Limit[Sum[Sin[Pi*k/n]/(n + 1/k), {k, 1, n}], n -> Infinity]


This sample calculates well in v8.0.4:

but not in v9.0.1 and v10.0 (tested on Cloud):

So this seems to be a backslide.

Any work-around?

There's no doubt that the answer for the infinite summation is $$2/\pi$$, this can be proved by squeeze theorem:

$$\because 0\leq1/k\leq1$$ $$\therefore \lim_{n\to \infty } \, \left(\sum _{k=1}^n \frac{\sin \left(\frac{\pi k}{n}\right)}{n+1}\right)\leq\lim_{n\to \infty } \, \left(\sum _{k=1}^n \frac{\sin \left(\frac{\pi k}{n}\right)}{\frac{1}{k}+n}\right)\leq\lim_{n\to \infty } \, \left(\sum _{k=1}^n \frac{\sin \left(\frac{\pi k}{n}\right)}{n}\right)$$ $$\lim_{n\to \infty } \, \frac{n \sum _{k=1}^n \frac{\sin \left(\frac{\pi k}{n}\right)}{n}}{n+1}\leq\lim_{n\to \infty } \, \left(\sum _{k=1}^n \frac{\sin \left(\frac{\pi k}{n}\right)}{\frac{1}{k}+n}\right)\leq\lim_{n\to \infty } \, \left(\sum _{k=1}^n \frac{\sin \left(\frac{\pi k}{n}\right)}{n}\right)$$ $$\because\lim_{n\to \infty } \, \left(\sum _{k=1}^n \frac{\sin \left(\frac{\pi k}{n}\right)}{n}\right)=\int_0^1 \sin (\pi x) \, dx=\frac{2}{\pi }$$ $$\lim_{n\to \infty } \, \frac{n}{n+1}=1$$ $$\therefore\lim_{n\to \infty } \, \left(\sum _{k=1}^n \frac{\sin \left(\frac{\pi k}{n}\right)}{\frac{1}{k}+n}\right)=\frac{2}{\pi }$$

• @Jinxed The sum is a Riemann Sum for Integrate[Sin[Pi x], {x, 0, 1}], which equals 2/Pi. – Michael E2 Apr 2 '15 at 13:07
• @xzczd Yes, that's quite nice, but I did not doubt the result. Note that the LHS is also a Riemann sum for the same integral if you sum $k=0$ to $n$ with $n+1$ intervals. – Michael E2 Apr 3 '15 at 2:47
• @MichaelE2 Yeah, I know, I just feel obligated to remind all the people who take part in the discussion for the correctness of v8 :) – xzczd Apr 3 '15 at 2:54
• By Laplace Transform: InverseLaplaceTransform[ Limit[Sum[ LaplaceTransform[Exp[I a*Pi*k/n]/(n + 1/k), a, s], {k, 1, n}] // FullSimplify, n -> Infinity, Assumptions -> s > 0], s, 1] // Im – Mariusz Iwaniuk Apr 2 '18 at 14:43
• Another workaround is to find the series at infinity. In[60]:= Series[ Sum[Sin[Pi*k/n]/(n + 1/k), {k, 1, n}], {n, \[Infinity], 0}] // Normal // Expand Out[60]= 2/\[Pi] – Chip Hurst Apr 2 '18 at 21:42

One possible workaround is to use the new in M12 function AsymptoticSum:

AsymptoticSum[Sin[Pi k/n]/(n + 1/k), {k, 1, n}, {n, ∞, 1}]


2/𝜋

Yesterday I found the approach below with Hold/ReleaseHold on v10.0.0 on Win8.1 achieves the same result as v8.0.4, namely, it gives a limit of $\frac{2}{\pi}$.

ReleaseHold@Limit[
Hold[
Sum[Sin[Pi*k/n]/(n + 1/k), {k, 1, n}]
], n -> Infinity]
(* 2/Pi *)


However, on v10.0.2 on Linux, this approach gives the result shown below...as does Wolfram Alpha. Also in a comment by Jinxed below, apparently this is the result in v10.1! Can anyone confirm the result on other operating systems or versions?

$Version (* 10.0 for Linux x86 (64-bit) (December 4, 2014) *) Limit[Sum[Sin[Pi*k/n]/(n + 1/k), {k, 1, n}], n -> Infinity]  Addendum Numerically, one can try this to investigate the behaviour of the function as n increases, and it looks like$\frac{2}{\pi}$is along the right lines. func = Compile[{{n, _Integer}}, 2/Pi - Sum[Sin[Pi*k/n]/(n + 1/k), {k, 1, n}] ]; data = Table[func[n], {n, 10000}]; ListLogPlot[data, Frame -> True, FrameLabel -> {"n", "func[n]"}]  On top of that, you can convert the LerchPhi mess into a function and plot the behaviour of that (include Chop to remove some small imaginary components). It's the same behaviour as the original sum: func2[n_] := 2/Pi - Chop@N@(-((1/(2*(-1 + E^((I*Pi)/n))*n^2))*((I*((-E^((I*Pi)/n))*n - E^((2*I*Pi)/n)*n + (E^(-((I*Pi)/n)))^(-1 + n)*n + E^((2*I*Pi)/n)*(E^((I*Pi)/n))^n*n - LerchPhi[E^(-((I*Pi)/n)), 1, 1 + 1/n] + E^((I*Pi)/n)* LerchPhi[E^(-((I*Pi)/n)), 1, 1 + 1/n] - (E^(-((I*Pi)/n)))^(-1 + n)* LerchPhi[E^(-((I*Pi)/n)), 1, 1 + 1/n + n] + (E^(-((I*Pi)/n)))^n* LerchPhi[E^(-((I*Pi)/n)), 1, 1 + 1/n + n] + E^((2*I*Pi)/n)*LerchPhi[E^((I*Pi)/n), 1, 1 + 1/n] - E^((3*I*Pi)/n)*LerchPhi[E^((I*Pi)/n), 1, 1 + 1/n] - E^((2*I*Pi)/n)*(E^((I*Pi)/n))^n*LerchPhi[E^((I*Pi)/n), 1, 1 + 1/n + n] + E^((3*I*Pi)/n)*(E^((I*Pi)/n))^n*LerchPhi[E^((I*Pi)/n), 1, 1 + 1/n + n]))/E^((I*Pi)/n))))?Q data2 = ParallelTable[func2[n], {n, 1, 1000, 5}]; ListLogPlot[Transpose[{Range[1, 1000, 5], data2}], Frame -> True, FrameLabel -> {"n", "func2[n]"}]  • My guess is that version 8 can't symbolically evaluate the Sum, which would have the same effect as using Hold/ReleaseHold in 9 and 10. – 2012rcampion Apr 1 '15 at 14:21 • @2012rcampion Interesting idea! – dr.blochwave Apr 1 '15 at 14:22 • @xzczd, can you try evaluating the sum by itself in v8? – 2012rcampion Apr 1 '15 at 14:24 • @blochwave, @xzczd: The Hold-ReleaseHold-approach does not work in 10.1! – Jinxed Apr 1 '15 at 19:27 • @Jinxed Oh! What does it give, out of interest? – dr.blochwave Apr 1 '15 at 20:12 ## Update Since you gave a good proof, that$2/\pi$is the correct solution, Mathematica is obviously failing at the task. The question is, why. ## Analysis of Mathematica's behavior First, the sum you gave is no Riemann sum: You defined the intervals as $$\Delta x_k=\frac{1}{n+1/k}$$ so$k/n$does not lie within the appropriate subinterval, e.g. for$n=10$,$k/n=\frac{1}{10}$(with$k=1$), while the interval width$1/(n+1/k)=\frac{1}{11}$is less, observe: NumberLinePlot@{Flatten@{FoldList[Plus, 0, Table[(10 + 1/k)^-1, {k, 1, 10}]], 1}, Table[k/10, {k, 1, 10}]}  So, this is no Riemann sum. Mathematica deals with Rieman sums easily: General Riemann-sum Riemann sum for$[a,b]$: riemann[f_,a_,b_,n_]:=With[{dx=(b-a)/n}, dx Sum[f[k dx], {k, 1, n}]] Limit[riemann[Sin[\[Pi] #]&, 0, 1, n], n->Infinity] (* 2/\[Pi] *)  ## Mathematica's behavior We have no Riemann sum here. Mathematica now seems to resort to LerchPhi and exponentialization, as soon as$k$and$n$appear together in the sum's divisor, possibly due to the Lerch zeta-function being defined as: $$L(\lambda ,n ,s)=\sum _{{k=0}}^{\infty }{\frac {\exp(2\,\pi \,i\,\lambda \,k)}{(k+n)^{s}}}$$ Now, Mathematica seems to find enough resemblance in this to begin transforming the original summand into a Lerch-based form, ending up with the complex expression in the question, for which it then cannot find a limit anymore. ## Conclusion: Solution? Having tried a number of documented summation methods in Sum, none of them prevent the expansion to Lerch. Since @blochwave's Hold and ReleaseHold does not work in 10.1 anymore, I am at a loss on how to actually have Mathematica find the limit of$2/\pi\$.

The solution would be to prevent Mathematica from trying LerchPhi at all - how this might be accomplished is however beyond me. :|

• +1 interesting work - I like it! The behaviour of the different versions of MMA w.r.t Hold/ReleaseHold is odd though - esp. on different OSs – dr.blochwave Apr 2 '15 at 14:52
• Aren't all the answers returned by the different Mathematica versions in fact equal? I can't check now, but I thought that's what I got. – Michael E2 Apr 2 '15 at 15:26
• @blochwave: What I wanted to say was: I am unable to manually evaluate the (long, LerchPhi) result myself. I just think, that the OP's original query does not immediately lend itself as being similar (or even equal) to the Riemann sum. – Jinxed Apr 2 '15 at 17:54
• I'm sorry but the answer given by v8 is undoubtedly correct. See my edit for the step-by step proof. @blochwave – xzczd Apr 3 '15 at 2:33
• @xzczd, @blochwave: I think I might have found a reason for LerchPhi showing up at all (see my update). Alas, now it seems utterly impossible to have Mathematica come up with the right result at all. – Jinxed Apr 3 '15 at 10:05