Chapter 10.4, Problem 62E

To determine

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

Explanation of Solution

Given:

The given series is $\mathrm{S}={\displaystyle \sum {\mathrm{a}}_{\mathrm{n}}}$ with no negative terms and ${\displaystyle \sum {\mathrm{d}}_{\mathrm{n}}}$ 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 ${\displaystyle \sum {\mathrm{d}}_{\mathrm{n}}}$ upto $\mathrm{K}\mathrm{t}\mathrm{h}$ term is,

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

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

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

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

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

To determine

To explain why it leads to contradiction if it is assumed that ${\displaystyle \sum {\mathrm{d}}_{\mathrm{n}}}$ is divergent.

Explanation of Solution

Given:

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

Calculation:

From part (a), if $\mathrm{S}={\displaystyle \sum {\mathrm{a}}_{\mathrm{n}}}$ is convergent, that would imply ${\displaystyle \sum {\mathrm{d}}_{\mathrm{n}}}$ is also convergent, which contradict the given information. Since, given ${\displaystyle \sum {\mathrm{d}}_{\mathrm{n}}}$ is divergent .