Sum of series

[Cereal]

How to calculate this sum:

 

[Steven G]

You solve it by using formulas you learned. What have you tried? Where are you stuck?

 

[Romsek]

Mathematica comes up with nothing for it. That doesn't mean it has no closed form expression but it does mean it's unlikely that it does.

 

[pka]

Mathematica comes up with nothing for it. That doesn't mean it has no closed form expression but it does mean it's unlikely that it does.

WolframAlpha finds convergence by comparison. But I be darn if I can find something to compare.
The standard reference in series also has it converging. I guess I just have been away too long.

P.S. It is very clear that \(\displaystyle \sum\limits_{n = 1}^\infty {\left( {1 - \sin ({n^{ - 1}})} \right)} \) diverges by the zero test.

 

[pka]

How to calculate this sum:
View attachment 15618

I was waiting, hoping, Cereal would post again. I was also looking an elementary comparison not knowing what s/he knows.
I gave up and applied the limit comparison test. The first on tried, \(\displaystyle \frac{1}{2^n}\). works beautifully.
To Cereal, do you understand that test?