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Trishann RussellWGU Student Number000779161Calculus III – C656Task 4Alternating Series Converge if:1.an>02.a1>a2>a3…..3.limn→∞an=0∑j=1∞(−1)j1√j+c=∑j=1∞(−1)j1√j+29where aj=¿f(j) = 1√(j+29)f(j) is positive if both numerator and denominator have the same sign. In 1√j+29the numerator is a positive constant. Therefore the denominator needs to be positive.(√(j+29)) =0 (√j+29)¿¿2=02(j+29)=0j=−29The denominator will be positive for all j¿-29 since we are only looking at [1, ∞) the denominator will always be greater than or equal to √(1+29)=√30 so both numerator and denominator will always be positive therefore f(j) will always be positive.If it is decreasing f’(j) must be negative between [1, ∞) f(j)=1√j+29=1(j+29)12=(j+29)−1