# 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. Apr 2, 2015 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. Apr 3, 2015 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 :) Apr 3, 2015 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 Apr 2, 2018 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] Apr 2, 2018 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. Apr 1, 2015 at 14:21 • @2012rcampion Interesting idea! Apr 1, 2015 at 14:22 • @xzczd, can you try evaluating the sum by itself in v8? Apr 1, 2015 at 14:24 • @blochwave, @xzczd: The Hold-ReleaseHold-approach does not work in 10.1! Apr 1, 2015 at 19:27 • @Jinxed Oh! What does it give, out of interest? Apr 1, 2015 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 Apr 2, 2015 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. Apr 2, 2015 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. Apr 2, 2015 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 Apr 3, 2015 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. Apr 3, 2015 at 10:05