# SumConverge does not give a valid result

I wanted to see if the function SumConverge worked for the following summation: $$\sum_{n=1}^{\infty}\frac{\sin^2(n)}{n^p}$$ We can clearly see that for $p>1$ the series converges and for $p\le1$ diverges by the comparison test. I tried:

SumConvergence[(Sin[n])^2/n^4, n]


However, I do not get an evaluation. Why so? I also tried $$\sum_{n=1}^{\infty}\left|\frac{\sin(n)}{n^p}\right|$$ Whose convergence or divergence can by found by Dirichlet's Test. However mathematica seems to not understand such input. I tried:

SumConvergence[Abs[(Sin[n])/n], n]


You need to give SumConvergence a little help:

SumConvergence[TrigReduce[Sin[n]^2/n^2], n]

True


previous post (may or may not be correct implementation)

SumConvergence[(Sin[n])^2/n^p, p] (* this gives condition for convergence *)
(* Abs[n] > 1 *)

SumConvergence[(Sin[Abs@n > 1])^2/n^2, n]
(*True*)

SumConvergence[(Sin[Abs@n > 1])^2/(n^-1), n]
(*False*)


new post

SumConvergence[(Sin[n])^2/n^p, p] (* this gives condition for convergence *)
(* Abs[n] > 1 *)


I think you should just use Sum and put VerifyConvergence -> True. I just read in the documentation that Sum automatically calls SumConvergence.

Sum[(Sin[n])^2/(n^4), {n, 1, \[Infinity]}, VerifyConvergence -> True]

(* 1/180 (\[Pi]^4 - 45 PolyLog[4, E^(-2 I)] - 45 PolyLog[4, E^(2 I)]) *)

Sum[(Sin[n])^2/(n^-1), {n, 1, \[Infinity]}, VerifyConvergence -> True]
(* Sum::div: Sum does not converge. >> *)

• Limit test does not prove if a sum is convergent. I want to use SumConvergence for p=4 – user372003 Mar 1 '17 at 21:04
• @user372003 does this suffice Sum[Sin[n]^2/n^4, {n, 1, \[Infinity]}]. the output is 1/180 (\[Pi]^4 - 45 PolyLog[4, E^(-2 I)] - 45 PolyLog[4, E^(2 I)]) – Ali Hashmi Mar 1 '17 at 21:19
• I specifically stated that I would prefer the use of SumConvergence instead of Sum. Although, I appreciate you trying to help. So the convergence of such sequence cannot be evaluated with SumConvergence? Also, if you try Sum[Sin[n]^2/n,{n,1,[Infinity]}] you don't get an an evaluation, why? – user372003 Mar 1 '17 at 21:24
• @user372003 can you check my answer again. if it makes sense now – Ali Hashmi Mar 1 '17 at 21:27
• Looks good for p>1 but does not work for p=1. Can you explain why you put Abs@n>1? Thank you! – user372003 Mar 2 '17 at 0:29