Integral Test — Question 4

Question 4 – Solution

Step 1: Verify the hypotheses.

Let $f(x)=x/(x^2+1)^2$ for $x\ge1$. It is continuous and positive. Differentiation gives $$f'(x)=\frac{1-3x^2}{(x^2+1)^3}<0\qquad(x\ge1),$$ so it decreases.

Step 2: Evaluate the integral.

With $u=x^2+1$ and $du=2x\,dx$, $$\int_1^{\infty}\frac{x}{(x^2+1)^2}\,dx =\frac12\int_2^{\infty}u^{-2}\,du=\frac14.$$ The integral converges, so the series converges.

Step 3: Bound the tail.

Since $$\int_A^{\infty}\frac{x}{(x^2+1)^2}\,dx=\frac1{2(A^2+1)},$$ the remainder satisfies $$\boxed{\frac1{2((N+1)^2+1)}\le R_N\le\frac1{2(N^2+1)}}.$$