Series 215 problem 03.mws

1. The Integral test for convergence.

The integral test is used to determine whether certain series converge. In order to use the integral test on the

series sum(a[n],n = 1 .. infinity) , it is necessary to find a continuous, positive, and decreasing function f which satisfies f(n) = a[n] .

For example, let a[n] = 1/(n*ln(n+1)) . Then a natural choice for the function f is f(x) = 1/(x*ln(x+1)) .

This function is continuous and positive for 1 <= x . To check that it is also decreasing , let us examine its derivative.

> f:=x->1/(x*ln(x+1));

f := proc (x) options operator, arrow; 1/(x*ln(x+1)...

> simplify(D(f)(x));


Note that every term in the fraction is positive when 1 < x , and the fraction is multiplied by -1, so the derivative of the function f is negative, which means the function is decreasing. Thus we can use this function in the integral test, because all the hypotheses of the integral test are satisfied. Let us compute the integral.

> int(f(x),x=1..infinity);evalf(%);

int(1/(x*ln(x+1)),x = 1 .. infinity)


Whatever Maple means by Float(infinity), we surely don't know, but Maple has not told us whether this integral converges, and it means that we will have to compare this integral to one which we can evaluate. Let g(x) = 1/((x+1)*ln(x+1)) . Then

it is easy to see that 0 < g(x) < f(x) on [ 1, infinity ). We can calculate Int(g(x),x = 0 .. infinity) using Maple:

> int(1/((x+1)*ln(x+1)),x=1..infinity);


But by the comparison test for integrals, it follows that Int(f(x),x = 0 .. infinity) diverges because Int(g(x),x = 0 .. infinity) diverges. From this result, we obtain that our original series also diverges.


For the series Sum(a[n],n = 1 .. infinity) below, do the following.

a) Find a function f which satisfies f(n) = a[n] .

b) Show that your function is positive, continuous, and decreasing.

c) Determine whether the integral of your function converges or diverges.

d) Determine whether the series converges or diverges.

1) Sum(ln(n)/(n^2),n = 3 .. infinity)

2) Sum(ln(n)/sqrt(n),n = 8 .. infinity)

Submission worksheet:


2. Another integral test for convergence.

The integral test is used to determine whether certain series converge. In order to use the integral test on a positive term series sum(a[n],n = 1 .. infinity) , it is necessary to find a continuous, positive, and decreasing function f which satisfies f(n) = a[n] . Consider the series, Sum(1/(n^4),n = 1 .. infinity) . This series is convergent since it is a p-series with p = 4 > 1, but let us argue carefully that it is convergent using the integral test. First, we choose f(x) = 1/(x^4) , and observe that f(n) = a[n] . Also for 1 <= x, 0 < f(x)

(in fact it is always positive valued, but we don't need this) and f is continuous on [1, infinity] . To show that f is decreasing we take its derivative and show that this is negative on [1, infinity] . We could get Maple to differentiate, but this is easier to do it by hand, so D(f)(x) = -4/(x^3) which is negative if 1 <= x . Therefore (and this is important), f satisfies the hypotheses of the integral test. So if the series above will converge if the improper integral, int(1/(x^4),x = 1 .. infinity) does, and diverge if the

improper integral diverges. A one line Maple code confirms what we already know.

> int(1/x^4,x=1..infinity);


Therefore the series sum(1/(n^4),n = 1 .. infinity) converges by the integral test.


(a) Show that the series sum(ln(n)^2/(n^2),n = 1 .. infinity) is convergent.

(b) Find an upper bound for the error in the approximation S[N] = sum(ln(n)^2/(n^2),n = 1 .. N) .

(c) What is the smallest value for N such the upper bound is less than 0.05 ?

(d) Find S[N] for this value of N .

(hint: you may want to use the comparison theorem for improper integrals)

Submission worksheet:


3. Ratio and root tests.

Let us consider the series whose n 'th term is given by a[n] = (2^n+5)/(3^n) . To apply the ratio test, we need to consider Limit(a[n+1]/a[n],n = infinity) . Let us define the n 'th term as a function of n and use Maple to compute this limit.

> a:=n->(2^n+5)/(3^n);

a := proc (n) options operator, arrow; (2^n+5)/(3^n...

> limit(a(n+1)/a(n),n = infinity);


Since this limit is less than 1, the series converges. We could also try the root test, which means we need to compute limit(a[n]^(1/n),n = infinity) .

> limit(a(n)^(1/n),n = infinity);


According to the root test, since 2/3<1, the series converges. Of course, we could also compute the sum of the series directly, and verify it converges this way. With Maple, we calculate

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

Sum((2^n+5)/(3^n),n = 0 .. infinity) = 21/2

We really don't need Maple to compute this sum, because the series can be broken up as a sum of two geometric series.

> Sum((2^n+5)/(3^n),n=0..infinity)=Sum((2/3)^n,n=0..infinity)+Sum(5*(1/3)^n,n=0..infinity);

Sum((2^n+5)/(3^n),n = 0 .. infinity) = Sum((2/3)^n,...

The first series converges to 1/(1-2/3) = 3 , while the second converges to 5/(1-1/3) = 15/2 , which explains the sum of the whole series.


For the following series, use Maple to apply the ratio and the root tests. Determine whether the series converges.

1) a[n] = n!/(n^n) .

2) a[n] = n!^n/((n^n)^2) .

Submission worksheet:


4. Absolute convergence, Ratio and root tests.

You have been introduced to the ideas of absolute and conditional convergence along with the ratio and root tests.


Determine if the following series are absolutely convergent, conditionally convergent or divergent.

If the series is convergent, use Maple to approximate the sum by finding the sum of the first 100 terms.

(a) Sum((-1)^(n-1)*sqrt(n)/(n+1),n = 1 .. infinity)

(b) Sum((-1)^n/(n*ln(n)),n = 2 .. infinity)

(c) Sum((-2)^n*n^2/(n+2)!,n = 1 .. infinity)

Submission worksheet: