Under what condition does the series converge, and what test should be used Thx

[Piking]

Under what condition does the series converge, and what test should be used

\(\displaystyle \displaystyle{\sum_{n\, =\, 1}^{\infty}}\, \left(\dfrac{1}{n^a}\, -\, \sin\left(\dfrac{1}{n^a}\right)\right)\)

...with \(\displaystyle a\, >\, 0.\)

 

[stapel]

Under what condition does the series converge, and what test should be used

\(\displaystyle \displaystyle{\sum_{n\, =\, 1}^{\infty}}\, \left(\dfrac{1}{n^a}\, -\, \sin\left(\dfrac{1}{n^a}\right)\right)\)

...with \(\displaystyle a\, >\, 0.\)

What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

Please be complete. Thank you!

 

[Piking]

I do not if my way is right! maybe tell me how you start?

 

[Ishuda]

I do not if my way is right! maybe tell me how you start?

As a hint, the way I would first approach this is to note that the expansion of the sin(x) is an alternating series and therefore it would be easy to get an lower and upper bound on the expression which was arbitrarily close to one another if a were large enough. Of course, there is no guarantee of success but I'm moderately certain that is the case without working it out.

Oh and I would also note that if a>1, both series converge individually.

 

[HallsofIvy]

I do not if my way is right! maybe tell me how you start?

I presume you mean "I do not know if my way is right". In that case, you should tell us what your way is and what answer you got.