# Solutions to worksheet problems

1. This series converges. Do fraction subtraction to see that it is equal to . You can do limit comparison of this series with .

2. This series also converges. You can compare it (not limit comparison) to .

3. This series diverges. Let . If we also let , then we have . By using the comparison test, we can say diverges when diverges.

So now we just need to check that diverges. For this, we can just use limit comparison with . This is a slightly different explanation than that given in class. If you understood the in-class version better, feel free to stick with it. Either type of answer will be acceptable on the midterm.

4. This problem is written incorrectly since the 's should have started with 2 instead of 1. By starting at 1, there is a division by zero in the first term. This is what lead to the confusion about how to apply the integral test in class.

Let me review what I said about the integral test in class. If you know is a positive and decreasing function, then you can always use the test to see if diverges or converges. Usually you try to evaluate the integral

but it turns out that you can choose any number between 1 and , and then try to integrate

instead. Your answer should always be the same. The reason you can not start the integration at 1 in this case is the division by zero in the first term.

Anyway, let's do the integration. We'll use 2 as the lower limit of integration. Make the substitution so that , and the integral becomes

The integral converges, and thus so does the series.

5. Use the ratio test to find the interval of convergence.

Take the limit as and we get the value . Remember that the ratio test gives convergence when the limit and divergence when . Thus the interval of convergence is , or any with (we don't know about the endpoints). The radius of convergence is 1.

so we can transform our function into and use the above expansion. The answer we get is

Now we should find the interval of convergence. Let's use the ratio test, ok?

and the limit is . Hence the interval is , or , which is the same as . The radius of convergence is .

and replace with . We get

Again use the ratio test to find the interval of convergence:

Hence the interval of convergence is , which is the same as , which is the same as . We can put this in the proper final format as . The radius of covergence is .

and replace with . This gives the series

9. For this one, you'll have to use the Taylor series theorem, which says that if a function has any power series representation at all (some functions have none), then it is given by:

So to find the proper coefficients, we'll need to find all the derivatives of . Let's get to work:

and all the higher derivatives are zero. Hence the Taylor series about is

. It is a finite series, so it converges everywhere.

10. First we need the series expansion for , which is

Now plug this into the fraction in the problem to get

that is, the limit is 1/6.

Calc 2 Homepage
Tyler Neylon
2003-04-06