Series 215 problem 02.mws

1. Summing a convergent series in Maple.

Let us consider the series whose k 'th term is given by a[k] = 1/((k+1)*(k+2)) . To use Maple to test for convergence, we can ask it to sum the series.

> Sum(1/((k+1)*(k+2)),k=1..infinity)=sum(1/((k+1)*(k+2)),k=1..infinity);

Sum(1/((k+1)*(k+2)),k = 1 .. infinity) = 1/2

One can ask how did Maple determine that the series converged to 1/2. Let us find a formula for the n 'th partial sum, using Maple's help.

> Sum(1/((k+1)*(k+2)),k=1..n)=sum(1/((k+1)*(k+2)),k=1..n);

Sum(1/((k+1)*(k+2)),k = 1 .. 482) = 241/484

Once we see the formula for this partial sum, it is very easy to see why the series converges to 1/2. But how could we have discovered this without Maple. Let us try to calculate the first few partial sums.

> for n from 1 to 5 do sum(1/((k+1)*(k+2)),k=1..n); od; n:='n';

1/6

1/4

3/10

1/3

5/14

n := 'n'

The Pattern is not immediately obvious, but if you change the denominators of the second and fourth partial sum to 8 and 12, then the pattern looks like 1/6, 2/8, 3/10, 4/12, 5/14, so that it is not hard to see that the numerator is just n , while the denominator is given by 2*n+4 . Then it is easy to use either L'Hopital's rule or Maple or some other simple method to see that

> Limit(n/(2*n+4),n=infinity)=1/2;

Limit(n/(2*n+4),n = infinity) = 1/2

Maple is not always capable of summing a series, even when we know what is going on. For example, take the series whose k 'th term is given by a[k] = 1/sqrt(k)-1/sqrt(k+1) . Let us see what Maple can do for us.

> sum(1/sqrt(k)-1/sqrt(k+1),k=1..n);

sum(1/(sqrt(k))-1/(sqrt(k+1)),k = 1 .. n)

Maple has just told us it does not know a closed form expression for the n 'th partial sum. What about the sum of the series?

> sum(1/sqrt(k)-1/sqrt(k+1),k=1..infinity);

sum(1/(sqrt(k))-1/(sqrt(k+1)),k = 1 .. infinity)

It does not appear that Maple understands how to sum this up exactly. But let us ask for a floating point estimate.

> evalf(sum(1/sqrt(k)-1/sqrt(k+1),k=1..infinity));

1.000000000

Finally, we get some useful information from Maple. Can we still use Maple to help us decide what is going on? Let us compute the first partial sums, to see what is going on.

> for n from 1 to 5 do sum(1/sqrt(k)-1/sqrt(k+1),k=1..n); od; n:='n';

1-1/2*sqrt(2)

1-1/3*sqrt(3)

1-1/4*sqrt(4)

1-1/5*sqrt(5)

1-1/6*sqrt(6)

n := 'n'

From these few terms, we see that the pattern is Sum(1/sqrt(k)-1/sqrt(k+1),k = 1 .. n) = 1-sqrt(n+1)... . It is easy to determine that the series converges to 1 from this formula. Moreover, this is a telescoping series, so it is easy to determine its convergence without any help from Maple.

Submission:

For the series whose k 'th term is given by a[k] as defined below, do the following:

a) Find the n 'th partial sum sum(a[k],k = 1 .. n) . Get as simple an expression for it as you can.

b) Find the limit of the n 'th partial sum limit(sum(a[k],k = 1 .. n),n = infinity) .

c) Verify the result in b) by having Maple try to compute the sum of the series sum(a[k],k = 1 .. infinity) .

d) Determine whether the series converges or diverges.

1) a[k] = (2*k+1)/(k^2*(k+1)^2)

2) a[k] = 1/(2^k)+(-1)^k/(5^k)

Submission worksheet:

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