Chapter 10.4, Problem 61E

To determine

To show the partial sums of ${\displaystyle \sum {\mathrm{a}}_{\mathrm{n}}}$ are bounded by ${\mathrm{a}}_1+\mathrm{.........}+{\mathrm{a}}_{\mathrm{N}}+{\displaystyle \sum _{\mathrm{n}=\mathrm{N}+1}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}}$

Explanation of Solution

Given:

The given series is $\mathrm{S}={\displaystyle \sum {\mathrm{a}}_{\mathrm{n}}}$ with no negative terms and ${\displaystyle \sum {\mathrm{c}}_{\mathrm{n}}}$ is the divergent series of non-negative terms.

Calculation:

The given series is $\mathrm{S}={\displaystyle \sum {\mathrm{a}}_{\mathrm{n}}}$ , the partial sum of the series up to $\mathrm{K}\mathrm{t}\mathrm{h}$ term is,

  ${\mathrm{S}}_{\mathrm{k}}={\mathrm{a}}_1+{\mathrm{a}}_2+\mathrm{.........}+{\mathrm{a}}_{\mathrm{N}}$

For $\mathrm{k}\le \mathrm{N}$ , ${\mathrm{S}}_{\mathrm{k}}={\mathrm{a}}_1+{\mathrm{a}}_2+\mathrm{.........}+{\mathrm{a}}_{\mathrm{k}}\le {\mathrm{a}}_1+{\mathrm{a}}_2+\mathrm{.........}+{\mathrm{a}}_{\mathrm{k}}+\mathrm{......}+{\mathrm{a}}_1+{\mathrm{a}}_2+\mathrm{.........}+{\mathrm{a}}_{\mathrm{N}}+{\displaystyle \sum _{\mathrm{n}=\mathrm{N}+1}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}}$

  ${\mathrm{S}}_{\mathrm{k}}={\mathrm{a}}_1+{\mathrm{a}}_2+\mathrm{.........}+{\mathrm{a}}_{\mathrm{k}}={\mathrm{a}}_1+{\mathrm{a}}_2+\mathrm{.........}+{\mathrm{a}}_{\mathrm{N}}+{\mathrm{a}}_{\mathrm{N}-1}+\mathrm{......}+{\mathrm{a}}_{\mathrm{k}}$

  ${\mathrm{S}}_{\mathrm{k}}={\mathrm{a}}_1+{\mathrm{a}}_2+\mathrm{.........}+{\mathrm{a}}_{\mathrm{N}}+{\mathrm{a}}_{\mathrm{N}-1}+\mathrm{....}+{\mathrm{a}}_{\mathrm{k}}\le {\mathrm{a}}_1+{\mathrm{a}}_2+\mathrm{.........}+{\mathrm{a}}_{\mathrm{N}}+{\mathrm{a}}_{\mathrm{N}-1}+\mathrm{......}+{\mathrm{a}}_{\mathrm{k}}$

  ${\mathrm{S}}_{\mathrm{k}}={\mathrm{S}}_{\mathrm{k}}={\mathrm{a}}_1+{\mathrm{a}}_2+\mathrm{.........}+{\mathrm{a}}_{\mathrm{N}}+{\mathrm{a}}_{\mathrm{N}-1}+\mathrm{....}+{\mathrm{a}}_{\mathrm{k}}\le {\mathrm{a}}_1+{\mathrm{a}}_2+\mathrm{.........}+{\mathrm{a}}_{\mathrm{N}}+{\mathrm{a}}_{\mathrm{N}-1}+\mathrm{......}+{\mathrm{a}}_{\mathrm{k}}+{\displaystyle \sum _{\mathrm{n}=\mathrm{N}+1}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}}$ Hence, the partial sum of the series is bound to ${\mathrm{a}}_1+\mathrm{.........}+{\mathrm{a}}_{\mathrm{N}}+{\displaystyle \sum _{\mathrm{n}=\mathrm{N}+1}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}}$

To determine

To explain why this shows $\mathrm{S}={\displaystyle \sum {\mathrm{a}}_{\mathrm{n}}}$ must converge.

Explanation of Solution

Given:

The given series is $\mathrm{S}={\displaystyle \sum {\mathrm{a}}_{\mathrm{n}}}$ with no negative terms and ${\displaystyle \sum {\mathrm{c}}_{\mathrm{n}}}$ is the divergent series of non-negative terms.

Calculation:

Since all the sum of $\mathrm{S}={\displaystyle \sum {\mathrm{a}}_{\mathrm{n}}}$ is non-negative. Therefore, the value of the partial sum of the series forms the non-decreasing sequence of the real numbers and since the series is bounded from part (a). Therefore, the series must be converges.