Question

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

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

Short answer

1. The sum of the series $\sum _{\mathrm{n}=1}^{\infty }\frac{\mathrm{n}}{{({\mathrm{n}}^2+1)}^2}=0.39711677$.
2. The sum of the series $\sum _{\mathrm{n}=2}^{\infty }\frac{\mathrm{In}\mathrm{n}}{{\mathrm{n}}^2}=0.93754825431$.
3. The sum of the series $\sum _{\mathrm{n}=1}^{\infty }\frac{1}{{\mathrm{n}}^{\mathrm{n}}}=1.29128599.$.

Step-by-step solution

Given Information

The given series are, i.e.,

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

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(n2-1)2

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

Hence, the sum of the series $\sum _{\mathrm{n}=1}^{\infty }\frac{\mathrm{n}}{{({\mathrm{n}}^2+1)}^2}=0.39711677$.

(b) Find the exact sum of the series ∑n=2∞In nn2

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

Hence, the sum of the series $\sum _{\mathrm{n}=2}^{\infty }\frac{\mathrm{In}\mathrm{n}}{{\mathrm{n}}^2}=0.93754825431$.

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

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

Hence, the sum of the series $\sum _{\mathrm{n}=1}^{\infty }\frac{1}{{\mathrm{n}}^{\mathrm{n}}}=1.29128599$.