Name: John Who is asking: Student Level: Secondary Question: I'm in a AP Calculus course and cannot solve the following series! And help would be greatly appreciated! Sincerely, John summation(n=1 to infinity)[n sin(1/(2n))]^{n} Hi John,First take the absolute value of each term and then apply the root test on this series. The nth root of n sin(1/(2n))^{n} is n sin(1/(2n)) and you can evaluate by making the substitution k = ^{1}/_{2n}. Cheers,Harley
