The series \(\sum {\frac{n^2-1}{n^2+1}x^2},x>0\) is

The correct answer is option 3.

Given: \(\sum {\dfrac{n^2-1}{n^2+1}x^2} , x > 0\)

Calculation:

⇒ \(\sum {\dfrac{n^2-1}{n^2+1}x^2} \)

Let's put x = 1

⇒ \(\sum {\dfrac{n^2-1}{n^2+1}(1^2)} \)  = \(\sum {\dfrac{n^2-1}{n^2+1}} \)

Now, put x = 2

⇒ \(\sum {\dfrac{n^2-1}{n^2+1}(2^2)} \) = \(\sum {4\dfrac{n^2-1}{n^2+1}} \)

So, the value of the function is increasing with the increasing value of x. 

Hence, the given function is divergent for the value of x ≥1.

Additional Information

• A series is said to be convergent, if it converges to a value as the terms of series tend to infinity.
• A series is said to be divergent, it does not converge to a value but keeps on either increasing or decreasing as the terms of series tend to infinity.