Hi! I'm pretty sure that the follow series diverges, but Wolfram Alpha claims that it converges. Why does Wolfram Alpha compute this wrong? $$\sum_{k=1}^{\infty}\frac{1}{k^{2-\cos(1/k)}}$$
The terms of the series are asymptotically equivalent to 1/k. Hence the series diverges with the same speed as the harmonic series:
1/k
Limit[k^(Cos[1/k] - 2)/k^-1, k -> Infinity] AsymptoticEqual[k^(Cos[1/k] - 2), 1/k, k -> Infinity]`enter code here`
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
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.
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
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!
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.
