Question

Find the sum of the following series ==> sum of n=1 to infinity 2/(4n^2-1)

Answer 1

We'll use: e = Σ [n=0→∞] 1/n! 

S = Σ [n=1→∞] (4n² - 1)/n! 
S = Σ [n=1→∞] (4.n(n-1) + 4n -1)/n! 
S = 4.Σ [n=1→∞] n(n-1)/n! + 4.Σ [n=1→∞] n/n! - Σ [n=1→∞] 1/n! 

Let's trim the initial number of the sum, using n(n-1)=0 for n=1 and 1/n! = 1 for n=0 
S = 4.Σ [n=2→∞] n(n-1)/n! + 4.Σ [n=1→∞] n/n! - Σ [n=0→∞] 1/n! + 1 

We may now simplify and start counting from zero: 
S = 4.Σ [n=2→∞] 1/(n-2)! + 4.Σ [n=1→∞] 1/(n-1)! - Σ [n=0→∞] 1/n! +1 
S = 4.Σ [p=0→∞] 1/p! + 4.Σ [q=0→∞] 1/q! - Σ [n=0→∞] 1/n! +1 
S = 7.Σ [n=0→∞] 1/n! + 1 
S = 7e + 1

Answer 2

Step-by-step explanation:

Find the sum of the series sigma n-infinity (4/(4n-3) (4n+1))