This applet illustrates the integral test for convergence of infinite series. Initially a sequence [math]a_n=f\left(n\right)[/math] where [math]f\left(x\right)[/math] is a positive, continuous, non-increasing function of [math]x[/math] is shown. The functions can be changed but make sure it is positive, continuous and non-increasing. A rectangle for each [math]a_n[/math] is also shown. The sum of the areas of the rectangles is equal to [math]\sum_1^Na_n[/math] and the start location of the rectangles can be shifted with the start slider. The number of terms can also be changed with the N slider.[br]Advancing to the next step shows [math]\int_1^Nf\left(x\right)dx[/math]. [br]With start = 0 and neglecting the area of [math]a_1[/math], how does the integral compare with the sum of the rectangle areas? With start = 1, how does the integral compare with the sum? [br]Advancing to the next step shows an equation representing the integral and sum relations and one more step shows the numerical values of the sums and integral.

By adding [math]a[br]_1[/math] to the left two terms the inequality can be rewritten as [math]\int_1^Nf\left(x\right)\le\sum_1^Na_n\le\int_1^Nf\left(x\right)+a_1[/math]. Then if [math]\lim_{N\to\infty}\int_1^Nf\left(x\right)\le M[/math] , the series is also bounded from the right side inequality. Also, if [math]\lim_{N\to\infty}\int_1^Nf\left(x\right)=\infty[/math], then from the left part of the inequality, the series is also unbounded. Or in other words, if the integral converges ( is bounded ) then the sum converges and if the integral diverges ( is unbounded ) the sum diverges.[br][br]Does the sum need to start at 1 for the integral test to be valid?[br][br]What would be required if [math]f\left(x\right)=\sin\left(\frac{4}{x}\right)[/math] for the same conclusions to be true?