Answer
(a) We show that $S = \mathop \sum \limits_{n = 1}^\infty \dfrac{n}{{{{\left( {{n^2} + 1} \right)}^2}}}$ converges.
(b) The approximation of $S$:
$0.397116 \le S \le 0.397117$
The maximum size of the error: ${10^{ - 6}}$
Work Step by Step
(a) For $n \gt 1$, we have
$0 \lt \dfrac{n}{{{{\left( {{n^2} + 1} \right)}^2}}} \lt \dfrac{n}{{{n^4}}}$
The larger series $\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{{n^3}}}$ converges (because $p = 3 \gt 1$, by the Convergence of $p$-Series).
Hence, by the Direct Comparison Test, the smaller series $\mathop \sum \limits_{n = 1}^\infty \dfrac{n}{{{{\left( {{n^2} + 1} \right)}^2}}}$ also converges.
(b) We have $S = \mathop \sum \limits_{n = 1}^\infty \dfrac{n}{{{{\left( {{n^2} + 1} \right)}^2}}}$.
Let $f\left( x \right) = \dfrac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}$.
Recall Eq. (4) in Exercise 85 of Section 11.3:
$\mathop \sum \limits_{n = 1}^M {a_n} + \mathop \smallint \limits_{M + 1}^\infty f\left( x \right){\rm{d}}x \le S \le \mathop \sum \limits_{n = 1}^{M + 1} {a_n} + \mathop \smallint \limits_{M + 1}^\infty f\left( x \right){\rm{d}}x$
With $M=99$, we get
$\mathop \sum \limits_{n = 1}^{99} \dfrac{n}{{{{\left( {{n^2} + 1} \right)}^2}}} + \mathop \smallint \limits_{100}^\infty \dfrac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}{\rm{d}}x \le S \le \mathop \sum \limits_{n = 1}^{100} \dfrac{n}{{{{\left( {{n^2} + 1} \right)}^2}}} + \mathop \smallint \limits_{100}^\infty \dfrac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}{\rm{d}}x$
We compute the following using a computer algebra system:
$\mathop \sum \limits_{n = 1}^{99} \dfrac{n}{{{{\left( {{n^2} + 1} \right)}^2}}} \approx 0.397066$, ${\ \ \ \ \ \ \ }$ $\mathop \sum \limits_{n = 1}^{100} \dfrac{n}{{{{\left( {{n^2} + 1} \right)}^2}}} \approx 0.397067$
$\mathop \smallint \limits_{100}^\infty \dfrac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}{\rm{d}}x \approx 0.000049995$
Thus,
$0.397066 + 0.000049995 \le S \le 0.397067 + 0.000049995$
$0.397116 \le S \le 0.397117$
From here, we obtain the maximum size of the error:
$0.397117 - 0.397116 = {10^{ - 6}}$
