Sum

For a positive real a and positive integer k, we have
k⁴ + a⁴ = k⁴ +2 k² a² +a⁴ - 2k²a² = (k² +a²)² -(√2 ka)² = (k² +√2 ka +a²) (k² - √2 ka +a²)
Using this the k-th term of the given sum could be written as
t_k = 12√2 a. 2√2ak (k² +√2 ka +a²) (k² - √2 ka +a²)
where I have taken a = 1/√2. The numerator can be written as
(k² +√2 ka +a²) - (k² - √2 ka +a²)
and so
t_k = 12√2 a(1k² - √2 ka +a² - 1k² + √2 ka +a²)
Plugging a =1/√2 we get
t_k = 12(1k² - k +(1/2) - 1k² + k +(1/2))
From here on its quite trivial to see that the sum telescopes to a value of
1 - 1n² + (n+1)²

4 k4 k ⁴ + 1

= 4 k( 1 + 2 k + 2 k ² ) ( 1 - 2 k + 2 k² )

= 11 - 2 k + 2 k ² - 11 + 2 k + 2 k ²

Hence , if we write down the sum , it'll look like -

11 - 15 + 15 - 113 + ......... + ( - 1 ) ^{n + 1} 11 + ( - 1 ) ⁿ 2 k + 2 k ²

Cancelling appropriate terms , we find the required sum to be as -

S ( n ) = 1 - ( - 1 ) ⁿ 11 + ( - 1 ) ⁿ 2 k + 2 k ²

It is rather easy to note that if " n → ∞ " , then " S ( n ) → 1 " .

You should note the use of the " Sophie - Germain " identity - For any two number " x " and " y " ,

x ⁴ + 4 y ⁴ = ( x ² - 2 x y + 2 y ² ) ( x ² + 2 x y + 2 y ² )