Answer
The partial sum ${S_{46}} \approx - 0.418452$ approximates the series $S = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{{n^3} + \sqrt n }} \approx - 0.418457$ to within an error of at most ${10^{ - 5}}$.
Work Step by Step
We have $S = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{{n^3} + \sqrt n }}$, where the positive terms are ${b_n} = \dfrac{1}{{{n^3} + \sqrt n }}$.
By Eq. (2) in Theorem 3 of Section 11.4:
$\left| {S - {S_n}} \right| \lt {b_{n + 1}}$
We determine $n$ so that
$\left| {S - {S_n}} \right| \lt {b_{n + 1}} \lt {10^{ - 5}}$
Thus,
$\dfrac{1}{{{{\left( {n + 1} \right)}^3} + \sqrt {n + 1} }} \lt {10^{ - 5}}$
${\left( {n + 1} \right)^3} + \sqrt {n + 1} \gt {10^5}$
We solve this inequality using a computer algebra system and obtain $n \gt 45.4148$.
So, we choose $N=46$.
Using a computer algebra system, we compute:
${S_{46}} = \mathop \sum \limits_{n = 1}^{46} \dfrac{{{{\left( { - 1} \right)}^n}}}{{{n^3} + \sqrt n }} \approx - 0.418452$, ${\ \ \ \ \ \ \ }$ $S = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{{n^3} + \sqrt n }} \approx - 0.418457$
$\left| {S - {S_{46}}} \right| \approx 4.969 \times {10^{ - 6}} \lt {10^{ - 5}}$
Thus, ${S_{46}} \approx - 0.418452$ approximates the series $S \approx - 0.418457$ to within an error of at most ${10^{ - 5}}$.
