The value of _ k=1 ^ (-1)^ k+1 ( k(k+1) k! ) is

Step-by-step solution

Evaluate _ k=1 ^ (-1)^ k+1 k(k+1) k! . Rewrite as _ k=1 ^ (-1)^ k+1 k+1 (k-1)! and split: _ k=1 ^ (-1)^ k+1 k (k-1)! + _ k=1 ^ (-1)^ k+1 1 (k-1)! . The first sum, after rewriting k (k-1)! = 1 (k-2)! + 1 (k-1)! for k 2 , evaluates to 0 using the exponential series _ j=0 ^ (-1)^j j! = e⁻¹ . The second sum with substitution j = k-1 gives _ j=0 ^ (-1)^j 1 j! = e⁻¹ = 1 e . Therefore, the total is 0 + 1 e = 1 e .