Question

f(x)= $\frac{1}{x+1}$ - $\frac{1}{x-1}$
Demonstrati ca $\lim_{n \to \infty} (f(2)+f(3)+f(4)+...+f(n) )$ = -3/2

Answer 1

$f(x) = \dfrac{1}{x+1}-\dfrac{1}{x-1} \\ \\ \lim\limits_{n\to \infty}\Big(f(2)+f(3)+f(4)+...+f(n)\Big) = \\ \\ = \lim\limits_{n\to \infty}\Big(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{n+1} - \dfrac{1}{1}-\dfrac{1}{2}-...-\dfrac{1}{n-1}\Big) = \\ \\ = \lim\limits_{n\to \infty}\Big(-1-\dfrac{1}{2}+\dfrac{1}{n}+\dfrac{1}{n+1}\Big) = -1-\dfrac{1}{2} = \boxed{-\dfrac{3}{2}}$