CHAPTER 8 - Infinite series

Section 8.1, page 619

Problems 35 and 36

We'll just calculate the limits:

>	limit(1-1/n,n=infinity), limit(n-1/n,n=infinity);

So the sequence in problem 35 converges to 1, and the one in problem 36 diverges.

Problem 55

>	a:=n->n^(1/n);

>	seq([n,evalf(a(n))],n=1..25);

Perhaps it's converging, let's see:

>	limit(a(n),n=infinity);

OK ... looks like we need to go pretty far out to get within 0.01:

>	fsolve(a(n)=1.01,n=25..50000);

So the error is not less than 0.01 until n=652. To get within 0.0001:

>	fsolve(a(n)=1.0001,n=1000..500000);

So we need 116,678 terms before we're within 0.0001.

Section 8.2, page 628

Problem 20

>	limit((sin(n))^2/2^n,n=infinity);

Problem 77

We want x = -cos(x). So we set

>	g:=x->evalf(-cos(x));

and then start with, say x=0.

>

g(0.0);

>

g(%);

>

g(%);

>

g(%);

>

g(%);

>

g(%);

>

g(%);

>

g(%);

So we seem to be converging to about -0.72.  Let's ask Maple:

>	fsolve(x+cos(x)=0,x=-1..0);

Section 8.3, page 638

Problem 39

>	Sum((exp(1)/Pi)^n,n=0..infinity)=sum((exp(1)/Pi)^n,n=0..infinity);

So this series converges.

Section 8.4, page 643

Problem 10

We'll integrate:

>	int(ln(x)/sqrt(x),x=2..infinity);

Since the integral diverges, so does the series.

Section 8.5, page 649

Problem 16

Since 1+ln(n)< for large n, as can be seen from this graph:

>	plot({1+ln(n),sqrt(n)},n=1..20,color=blue,thickness=2);

>	evalf(sqrt(20)),evalf(1+ln(20));

we see that   But

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

So both series diverge.

Section 8.6, page 654

Problem 17

Ratio test:

>	a:=n->(n+1)*(n+2)/n!;

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

So the series converges.

Section 8.7, page 661

Problem 26

Alternating series test:

>	limit(1/(n*ln(n)),n=infinity);

And the denominator is increasing, so the fraction is decreasing. So  converges at least. To check absolute convergence, we'll use the integral test:

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

This diverges, so the original alternating series converges conditionally.

Section 8.8, page 671

Problem 17

First use the ratio test:

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

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

>	solve(abs(%)<1,x);

So the series converges for x between -8 and 2. What about the endpoints?

>	simplify(subs(x=-8,a(n)));

>	simplify(subs(x=2,a(n)));

Since these don't go to zero (nth term test), the series diverges for x=-8 and x=2. So the interval of convergence is (-8,2) -- the series converges absolutely for all those values of x.

Section 8.9, page 677

Problem 34

We'll find an expansion of higher order, just to show we can:

>	taylor(exp(sin(x)),x=0,7);

The linear approximation is thus 1+x, and the quadratic approximation is .

Section 8.10, page 686

Problem 21

If |x|<0.001, and since the Maclaurin series for sin(x):

>	taylor(sin(x),x=0,8);

is alternating, the error is less than the first omitted term. So the error is less than

>

0.001/6;

i.e., less than 0.00017.

Section 8.11, page 697

Problem 4

>	taylor(sqrt(1+x),x=0,5);

Problem 44

>	taylor(Int(t^2*exp(-t^2),t=0..x),x=0,10);

This is an alternating series, so to get error less than 0.001 for  |x|<1, we need to find the first term with denominator greater than 1000:

>	taylor(Int(t^2*exp(-t^2),t=0..x),x=0,15);

So going up through the  term will do.