Calculus II

Lesson 17: Convergence of Series

Definition: A series consists of two sequences: The sequence of terms of the series and the sequence of partial sums of the series. If the sequence of partial sums converges to a limit L, then the series is said to converge to L and we write .

Suppose you have established somehow, either directly or by some test, that a series converges to a number L. How do we calculate this number to any specified accuracy?

Not surprizingly, Maple can sum a lot of series already. For example, the sum of a geometric series is easy for Maple to compute.

> restart;

> Sum(a*r^n,n=0..infinity)=sum(a*r^n,n=0..infinity);

Maple also knows that the harmonic series diverges to infinity.

> sum(1/n,n=1..infinity);

It also knows how to compute the sums of the various convergent p-series. For example, the 3-series sums to

> Sum(1/n^3,n=1..infinity)=sum(1/n^3,n=1..infinity);

The Riemann Zeta function is defined for all p > 1 to give the sum of the p-series.

So, for example, the sum of the 3 series is

> Zeta(3)=Zeta(3.);

If the series converges fast enough you can look at the sequence of partial sums and get the desired accuracy.

Lets see how fast the 3-series converges to Zeta(3).

> for n from 10 by 100 to 300 do Sum(1/i^3,i=1..n)=evalf(sum(1/i^3,i=1..n)) od;

Well, the sum of the first 100 terms is accurate to 4 significant figures.

You can't decide for sure by looking at first few partial sums of a series that the series converges. For example, look at a few partial sums of the harmonic series.

> seq(evalf(sum(1/i,i=1..100*n)),n=1..5); n:='n':

Hmmm. You can't really tell by looking at these that the harmonic series doesn't converge.

>

Problems

Exercises: In each of the problems below, determine whether the series converges or diverges. Give a reason in each case. For the convergent series, get an estimate correct to 2 decimal places of the sum of the series using psums or some other word of your own devising. You can check with sum to see if Maple can sum it.

This series converges by comparison with the p-series . Each partial sum is bounded above by and the partial sums form an increasing sequence, so we know the sequence of partial sums converge. Checking to see what is programmed into Maple,

> sum(1/(3+2*n)^2,n=1..infinity)=
evalf(sum(1/(3+2*n)^2,n=1..infinity));

we get an exact sum. Check a few partial sums .

> seq(evalf(sum(1/(3+2*i)^2,i=1..100*n)),
n=1..5); n:='n':

The function is a decreasing for n > 1 (Take the derivative), so we can use the integral test on this one.

> int(1/(n*(ln (n))^2),n=1..infinity);

Since the integral diverges, the series diverges.

This series diverges, since it is a p-series with p < 1

This series converges by comparison with .

> evalf(int(( n^5+4*n^3+1)/
(2*n^9+n^4+2 ),n=1..infinity));

This series converges by comparison with the 4-series.

> evalf(sum(sin(1/(n^4)),n=1..infinity));