Evaluating series Evaluate the following infinite series or state that the series diverges.
STEP_BY_STEP SOLUTION Step-1 Definition ; A series is said to be convergent if it approaches some limit .Formally , the infinite series a n is convergent if the sequence of partial sums n=1 n S n a is convergent . Conversely , a series is divergent if the k=1 ksequence ofpartial sums is divergent.NOTE : The terms grow without bound , so the sequence does not converge. If U k and V k are convergent series , then ( U + k k ) and ( U - Vk k ) areconvergent . If C = / 0 , then U k and V k both converge or both diverge . Step-2 2 3a + ar + ar +ar +..................... is an infinite geometric progression.Here the first term is ‘a’ , and the common ratio is ‘r’ n If |r| < 1 , then the sum of the series ar = a ……………(1) r=0 1 r rNOTE ; The infinite geometric series x converges if |x| < 1 , and diverges if |x| 1 r =0 Step-3\n The given sequence is S = 1 = ( - 1 1 ) k k(k+1) k (k+1) k =1 k =1 n Then S n ( -1 1 ) k =1 k (k+1) 1 1 1 1 1 1 1 = ( 1 - 2+ 2 - 3 + ………+ n1 - n + n - n+1 ) = ( 1 - 1 ) n+1 1 Therefore , S =n1 - n+1Now , we have to find the value of lim S = limn( 1 - 1 ) n n n+1 = lim ( 1 - n(1+ 1/n)). n = 1 , since as n , then 0. n 1 Therefore , k(k+1) = 1. k =1 From the step(1) , step(2) , it is clear that the given infinite series is convergent . Becausethe value approaches to ‘1’ (finite value). 1 Therefore , k(k+1) is convergent series. k =1