Question

By computer or tables, find the exact sum of each of the following series.

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{n}}{{\mathbf{(}\mathbf{4}{\mathbf{n}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}} \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{n}}^{\mathbf{3}}}{\mathbf{n}\mathbf{!}} \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{{\mathbf{3}}^{\mathbf{n}}} \end{gathered}$$

Short answer

1. The sum of the series$\sum _{n=1}^{\infty }\frac{n}{{(4n^2-1)}^2}=\frac{1}{8}$.
2. The sum of the series$\sum _{n=1}^{\infty }\frac{n^3}{n!}=13.5914091422=5e$.
3. The sum of the series $\sum _{n=1}^{\infty }\frac{n(n+1)}{3^n}=\frac{9}{4}$.

Step-by-step solution

Given Information

The given series are, i.e.,

$$\begin{gathered} \left(a\right)\sum _{n=1}^{\infty }\frac{n}{{(4n^2-1)}^2} \\ \left(b\right)\sum _{n=1}^{\infty }\frac{n^3}{n!} \\ \left(c\right)\sum _{n=1}^{\infty }\frac{n(n+1)}{3^n} \end{gathered}$$

Definition of Sum of a Series

The total of all the terms in a sequence can be solved as the sum of a series.

(a) Find the exact sum of the series ∑n=1∞n(4n2-1)2

Use the MATLAB tool to calculate the exact sum of the series,

Hence, the sum of the series $\sum _{n=1}^{\infty }\frac{n}{{(4n^2-1)}^2}=\frac{1}{8}$.

(b) Find the exact sum of the series ∑n=1∞n3n!

Use the MATLAB tool to calculate the exact sum of the series,

Hence, the sum of the series $\sum _{n=1}^{\infty }\frac{n^3}{n!}=13.5914091422=5e$.

(c) Find the exact sum of the series ∑n=1∞n(n+1)3n

Use the MATLAB tool to calculate the exact sum of the series $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{{\mathbf{3}}^{\mathbf{n}}}$,

Hence, the sum of the series $\sum _{n=1}^{\infty }\frac{n(n+1)}{3^n}=\frac{9}{4}$.