1.

(a) The similar problem is a geometric series with r = sin(x). So we set our r between -1 and 1 to get -1 < sin(x) < 1. That is, whenever sin(x) is not -1 or 1 the series converges. So the only problem is when sin(x) = 1 (this happens whenever x = pi/2 + 2*pi*n) or when sin(x) = -1 (this happens whenever x = - pi/2 + 2*pi*n). In fact, if you plug any of these x-values in, you either get an infinite sum of 1's (which diverges to infinity) or you get an infinite sum like -1 + 1 - 1 + 1 - 1 etc., which diverges because the sums don't get close to any one specific value. So the answer is: All x-value except those for which x = pi/2 + 2*pi*n or for which x = -pi/2 + 2*pi*n.
(b) The series is an infinite geometric series with r = sin(2); since sin(2) is strictly between -1 and 1, the series converges with sum sin(2)/(1-sin(2))

2.

(a) Take three derivatives of ln(x), plux in x = 1, and get (x-1)³/3 - (x-1)² + x - 1.
(b) The seventh derivative of ln(x) is 720x^-7. By graphing, we can see that the biggest this derivative gets (in absolute value!) is 720. So in the formula Error < = |M(x-1)⁷/720|, we get that Error < |0.5⁷| = 0.0078125.

3.

To find a power series for x²/(1-x)², we either need to start taking derivatives and do it the long way, OR compare this to our short list of series we already know, and ask ourselves what we would have to do to the known series to get the one we have. In this case, the series being asked for only looks like 1/(1-x), and the only nice thing I can think of to do to get that square term on the bottom is take the derivative. So we start with

and take the derivative of both sides to get

then multiply both sides by -x² to get

.

4.

(a). This series alternates, so if we can show that the terms are always getting smaller, and whose limit is 0, the alternating series test says that the series converge (this is the dance that stays in the room!) The limit does equal 0, and the terms do constantly get smaller, as can be seen by rewriting as follows:

so that you can see now that the bottom keeps getting bigger as n increases, and since the bottom increases without bound, the whole fraction gets arbitrarily small (the limit goes to 0!)

(b) Use the integral test; show that the integral diverges, then the series which is bigger, also diverges.

(c) Compare this series to the series whose nth term is given by a_n = 1/(n + n) = 1/(2n). Since this new series diverges (it's half the harmonic series, and half of infinite is still infinite), our series (which is bigger since it's denominator is smaller) must also diverge.

5.

Again, we're thinking of our function, in this case 1/(x²+1) as a sum of sines and cosines:

To find the constant coefficient, just integrate both sides from - pi to +pi. The integrals of all the trig functions over this period are 0 (for each, there's just as much on top of the x-axis as there is underneath.) So the integral on the right hand side is just 2*pi*a₀. The integral of the left-hand side is arctan(pi) - arctan(-pi). So a₀ = (arctan(pi) - arctan(-pi))/(2*pi), which is approximately 0.40191 (to 5 decimal places).

6.

We have to solve the equation

which we solve graphically and get a value of about i = 0.1429 (so about 14.3%).