Chapter 10.5, Problem 52E

To determine

The proof of the integral test for N=1.

Answer to Problem 52E

If the integral converges, then the second inequality puts an upper bound on the partial sums of the series.

Explanation of Solution

Given information:

The Integral Test (Theorem 10) for N = 1

  

Formula used:

  ${\displaystyle {\int }_1^{n+1}f\left(x\right)}dx<a_1+a_2+\mathrm{....}+a_n<a_1+{\displaystyle {\int }_1^{n}f\left(x\right)dx}$

Calculation:

For N=1 and $n\ge 1$ , then the sum $a_1+a_2+\mathrm{....}+a_n$ provides an upper bound for ${\displaystyle {\int }_1^{n+1}f\left(x\right)}dx$ and the sum $a_2+a_3+\mathrm{....}+a_n$ provides a lower bound for ${\displaystyle {\int }_1^{n}f\left(x\right)}dx$ , so ${\displaystyle {\int }_1^{n+1}f\left(x\right)}dx<a_1+a_2+\mathrm{....}+a_n<a_1+{\displaystyle {\int }_1^{n}f\left(x\right)dx}$

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 of the series, and since they are a non-decreasing sequence, they must converge to a finite sum for the series.

Conclusion:

If the integral converges, then the second inequality puts an upper bound on the partial sums of the series