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# Series is unexpectedly convergent by Wolfram|Alpha

Posted 2 years ago
 Hi! I'm pretty sure that the follow series diverges, but Wolfram Alpha claims that it converges. Why does Wolfram Alpha compute this wrong? $$\sum_{k=1}^{\infty}\frac{1}{k^{2-\cos(1/k)}}$$
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Posted 2 years ago
 The terms of the series are asymptotically equivalent to 1/k. Hence the series diverges with the same speed as the harmonic series: Limit[k^(Cos[1/k] - 2)/k^-1, k -> Infinity] AsymptoticEqual[k^(Cos[1/k] - 2), 1/k, k -> Infinity]enter code here 
Posted 2 years ago
 Hello folkx :) Wolfram|Alpha is RIGHT, this series converges. Here is the proof by a friend of mine (Jean-Pierre Delgado) : Well it is in French but so simple to read :) Have fun and keep cool :) Jean-Michel Collard
Posted 2 years ago
 The steps are correct but the conclusion is wrong. The last step means the series uniformly convergent to a divergent series. So the series is divergent. Simply speaking anything parallel to $1/x$ on log log plot or can be squeezed between two parallel lines are divergent. $k^{-1.2}$ is close but eventually it will go across any parallel line. Convergence test is always tricky topic if the result is close to one. Handle with care.
Posted 2 years ago
 Hello Shengui and others, Yes the proof USING Mathematica is clear enough and is absolutely right. But what I expected is a formal proof of the divergence withOUT using Mathematica. I just need a clue to start to prove it with pen and paper. Thank you for the very high quality of your post!Jean-Michel Collard
Posted 2 years ago
 Thank you Jim and Shenghui! Then we agree on that the sum in fact is divergent! Shenghui, thanks for reporting the issue to the Wolfram|Alpha math team. Could you please let me know if/when they find out what went wrong and how they were able to fix the issue? Thanks in advance!
Posted 2 years ago
 Looks like performance issue.Probably at the beginning Wolfram|Alpha looks for analytical solution if not found then use a numerically approximation.To quick calculation on web page Wolfram|Alpha use 200 terms but not very big number.Probably,if the change value of the sum is small, it concludes that the sum is convergent.
Posted 2 years ago
 This series should be divergent and it can be checked with Asymptotic function. I have send a feedback about this observation to the W|A math team. Thanks.
Posted 2 years ago
 If 200 is a close approximation to $\infty$, then Sum[k^(Cos[1/k] - 2), {k, 1, 200}] // N = 5.78299.But if one plots the partial sum vs the number of terms, one sees the following: t = Table[{n, Sum[k^(Cos[1/k] - 2), {k, 1, n}]}, {n, 1, 5000, 100}]; ListLogLinearPlot[Table[{n, Sum[k^(Cos[1/k] - 2), {k, 1, n}]}, {n, 1, 5000, 100}], PlotRange -> All] So I think you're correct in that the sum does not converge. But I don't know why Wolfram Alpha gives the answer it does.