Question

Estimate the Error if$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{x}}^{\mathbf{n}}}{{\mathbf{n}}^{\mathbf{3}}}$is approximated by the sum of its first three terms for$\left|x\right|\mathbf{<}\frac{\mathbf{1}}{\mathbf{2}}$.

Short answer

The Error is 0.001953.

Step-by-step solution

Given Information

The function is $\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^3}$.

Definition of the Error of the Series

The calculated Series is Underestimate if the Error is less than zero. The calculated Series is Overestimate if the Error is more than zero.

Verify the statement

$\mathrm{The}\mathrm{function}\mathrm{is}:$

$\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^3}$

The formula states that:

$\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^3}=x+\frac{x^2}{2^3}+\frac{x^3}{2^3}+\frac{x^4}{2^3}+....$

The theorem states that:

$$\begin{gathered} \mathrm{If}\mathrm{S}=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\mathrm{coverages}\mathrm{for}\left|\mathrm{x}\right|<1,\mathrm{and}\mathrm{if}\left|{\mathrm{a}}_{\mathrm{n}+1}\right|<\left|{\mathrm{a}}_{\mathrm{n}}\right|\mathrm{for}\mathrm{n}>\mathrm{N},\mathrm{then} \\ \left|\mathrm{S}-\sum _{\mathrm{n}=0}^{\mathrm{N}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\right|<\frac{\left|{\mathrm{a}}_{\mathrm{N}+1}{\mathrm{x}}^{\mathrm{N}+1}\right|}{\left(1-\left|\mathrm{x}\right|\right)} \\ \left|\mathrm{S}-\sum _{\mathrm{n}=0}^{\mathrm{N}}{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\right|<\frac{\left|{\mathrm{a}}_{\mathrm{N}+1}{\mathrm{x}}^{\mathrm{N}+1}\right|}{\left(1-\left|\mathrm{x}\right|\right)} \\ \left|\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^3}-\left(\mathrm{x}+\frac{{\mathrm{x}}^2}{2^3}+\frac{{\mathrm{x}}^3}{3^3}\right)\right|<\frac{\left|{\mathrm{a}}_{\mathrm{N}+1}{\mathrm{x}}^{\mathrm{N}+1}\right|}{\left(1-\left|\mathrm{x}\right|\right)} \\ \left|\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^3}-\left(\mathrm{x}+\frac{{\mathrm{x}}^2}{2^3}+\frac{{\mathrm{x}}^3}{3^3}\right)\right|<\frac{\left|{\displaystyle \frac{1}{4^3}}.{\displaystyle \frac{1}{2^4}}\right|}{1-\left|{\displaystyle \frac{1}{2}}\right|} \\ \left|\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^3}-\left(\mathrm{x}+\frac{{\mathrm{x}}^2}{2^3}+\frac{{\mathrm{x}}^3}{3^3}\right)\right|<\frac{0.000976}{0.5} \\ \mathrm{Solve}\mathrm{further},\mathrm{and}\mathrm{we}\mathrm{get}, \\ \left|\sum _{\mathrm{n}=1}^{\infty }\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^3}-\left(\mathrm{x}+\frac{{\mathrm{x}}^2}{2^3}+\frac{{\mathrm{x}}^3}{3^3}\right)\right|<0.001953 \\ \mathrm{Error}\approx 0.001953 \\ \mathrm{Hence},\mathrm{the}\mathrm{Error}\mathrm{is}0.001953. \end{gathered}$$