# Series is unexpectedly convergent by Wolfram|Alpha

Posted 1 month ago
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 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 1 month 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.
Posted 1 month 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 1 month 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 1 month 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 1 month 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 1 month 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.
 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