Chapter 9.5, Problem 53E

To determine

To state: Why the same conclusions about convergence can be drawn if the picture of an arbitrary N is relabeled.

Answer to Problem 53E

If the integral diverges, it must go to infinity, and the first inequality forces the partial sums of the series to go to infinity as well, so the series is divergent.

Explanation of Solution

Given information:

The given statement says to explain the reason about why the same conclusions about convergence can be drawn if the picture of an arbitrary N is relabeled.

The graphs are drawn for an arbitrary N as follows:

  

  

Comparing areas in the figures, it must be for all $n\ge N$ , the sum $a_N+a_{N+1}+\cdots +a_n$ provides an upper bound for ${\displaystyle \underset{N}{\overset{n+1}{\int }}f\left(x\right)dx}$ and the sum $a_{N+1}+a_{N+2}+\cdots +a_n$ provides a lower bound for ${\displaystyle \underset{N}{\overset{n}{\int }}f\left(x\right)dx}$, so it can be written as:

  ${\displaystyle \underset{N}{\overset{n+1}{\int }}f\left(x\right)dx}<a_N+\cdots +a_n<a_N+{\displaystyle \underset{N}{\overset{n}{\int }}f\left(x\right)dx}$

If the integral diverges, it must go to infinity, and the first inequality forces the partial sums of the series to go to infinity as well, so the series is divergent.

If the integral converges, then the second inequality puts an upper bound on the partial sums to the series, and since they are a no decreasing sequence, they must converge to a finite sum for the series.