Integral Test and p-Series

I.  Quiz

II.  Homework

III.  The Integral Test

Consider a series sum a_n such that a_n > 0 and a_n > a_n+1

We can plot the points (n,a_n) on a graph and construct rectangles whose bases are of length 1 and whose heights are of length a_n.  If we can find a continuous function f(x) such that f(n) = a_n, then notice that the area of these rectangles is an upper Reimann sum for the area under the graph of f(x).  Hence int from 1 to infinity of f(x) dx < sum from 1 to infinity of a_n .  Similarly, if we investigate the lower Reimann sum we see that

int from 1 to infinity f(x) dx > sum from 2 to infinity of a_n.

 This proves the following theorem:

Theorem:  The Integral Test

Let f(x) be a positive  continuous function that is eventually decreasing and let f(n) = a_n

Then

sum a_n converges if and only if int from 1 to infinity converges.

 Note the contrapositive:

sum a_n diverges if and only if int from 1 to infinity diverges.

Example:

A)  Consider the series:  sum 1/n²

We use the Integral Test:

Let f(x) = 1/x²  

Hence by the Integral Test sum 1/n² converges.

What does it converge to?  We use a calculator:  Go to MATH MISC then put

sum seq(1/x²,x,1,100,1) to get 1.635...

 B)  Consider the series sum 1/sqrt(n)

We use the Integral Test:

Let f(x) = 1/sqrt(x) 

int from 1 to infinity of  1/sqrt(x) dx = lim m -> infinity 2sqrt(x) from 1 to infinity = infinity.  Hence by the Integral Test sum 1/sqrt(n) diverges.

Note that if we use the calculator, we get

sum seq(1/sqrt(x),x,1,100,1) = 18.59

Hence, you cannot tell from the calculator whether it converges or diverges.

Theorem:  P-Series Test

Consider the series sum 1/n^p  

If p > 1 then the series converges

If 0 < p < 1 then the series diverges

Proof:

We use the integral test with the function f(x) = 1/x^p

For p not 1, lim int 1/x^p dx = lim int x^-p dx = lim x^-p+1/-p + 1 Note that this limit converges if

-p + 1 < 0 or p > 1.  The limit diverges for p < 1

For p = 1 we have the harmonic series

sum 1/n and the integral test gives:

lim int 1/x dx = lim log x = infinity.

Hence the harmonic series diverges.