Solutions to worksheet problems

1. This series converges. Do fraction subtraction to see that it is equal to $\sum\frac{2}{n(n+2)}$. You can do limit comparison of this series with $\sum\frac{1}{n^2}$.

2. This series also converges. You can compare it (not limit comparison) to $\sum\frac{1}{n^2}$.

3. This series diverges. Let $a_n=\frac{1+1/n}{\sqrt{n(n+2)}}$. If we also let $b_n=\frac{1}{\sqrt{n(n+2)}}$, then we have $a_n\ge b_n$. By using the comparison test, we can say $\sum a_n$ diverges when $\sum b_n$ diverges.

So now we just need to check that $\sum b_n$ diverges. For this, we can just use limit comparison with $1/n^2$. 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 $n$'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 $f(n)$ is a positive and decreasing function, then you can always use the test to see if $\sum_n^\infty f(n)$ diverges or converges. Usually you try to evaluate the integral

$$\int_1^\infty f(x)\,dx,$$

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

$$\int_a^\infty f(x)\,dx$$

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 $u=\log x$ so that $du = dx/x$, and the integral becomes

$$\int_{\log 2}^\infty \frac{du}{u^3} = \left[\frac{1}{-2u^2}\right]_{\log 2}^\infty = \frac{1}{2\log^2 2} < \infty.$$

The integral converges, and thus so does the series.

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

$$\left\vert\frac{a_{n+1}}{a_n}\right\vert = \left\vert\frac{(n+1)^2x^{n+1}}{n^2x^n}\right\vert = \left\vert\frac{(n+1)^2x}{n^2}\right\vert.$$

Take the limit as $n\to\infty$ and we get the value $\vert x\vert$. Remember that the ratio test gives convergence when the limit $L<1$ and divergence when $L>1$. Thus the interval of convergence is $\vert x\vert<1$, or any $x$ with $-1<x<1$ (we don't know about the endpoints). The radius of convergence is 1.

6. We already know that

$$-\log(1-x) = \int\frac{1}{1-x}dx = \int (1+x+x^2+\ldots) = x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots,$$

so we can transform our function $f(x) = \log\left( \frac{1}{1+2x}\right)$ into $f(x)=-\log(1-(-2x))$ and use the above expansion. The answer we get is

$$f(x) = -2x + \frac{4x^2}{2} - \frac{8x^3}{3} +\ldots = \sum_{n=1}^\infty\frac{(-2x)^n}{n}.$$

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

$$\left\vert\frac{a_{n+1}}{a_n}\right\vert = \left\vert\frac{(-2x)^{n+1}n}{(n+1)(-2x)^n}\right\vert = \left\vert\frac{2xn}{n+1}\right\vert$$

and the limit is $\vert 2x\vert$. Hence the interval is $\vert 2x\vert<1$, or $-1<2x<1$, which is the same as $-1/2<x<1/2$. The radius of convergence is $1/2$.

7. Start with

$$\frac{1}{1-x} = 1 + x +x^2 + \ldots$$

and replace $x$ with $-2x^2$. We get

$$f(x) = 1 - 2x^2 + 4x^4 \mp\ldots = \sum_{n=0}^\infty (-1)^n2^nx^{2n}$$

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

$$\left\vert\frac{a_{n+1}}{a_n} \right\vert = \left\vert\frac{2^{n+1}x^{2n+2}}{2^nx^{2n}}\right\vert = \vert 2x^2\vert.$$

Hence the interval of convergence is $\vert 2x^2\vert=2\vert x\vert^2<1$, which is the same as $\vert x\vert^2<1/2$, which is the same as $\vert x\vert<1/\sqrt 2$. We can put this in the proper final format as $-1/\sqrt 2<x <1/\sqrt 2$. The radius of covergence is $1/\sqrt 2$.

8. Start with the Taylor series for $e^x$, which is

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots,$$

and replace $x$ with $-x^3$. This gives the series

$$f(x) = 1 - x^3 + \frac{x^6}{2!} - \frac{x^9}{3!} \pm \ldots$$

which is the answer.

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

$$f(x) = \sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n.$$

So to find the proper coefficients, we'll need to find all the derivatives of $f(x) = x^4$. Let's get to work:

$$f'(x) = 4x^3$$

$$f''(x) = 12x^2$$

$$f^{(3)}(x) = 24x$$

$$f^{(4)}(x) = 24$$

and all the higher derivatives are zero. Hence the Taylor series about $a=2$ is

$$f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f^{(3)}(a)}{3!}(x-a)^3 + \frac{f^{(4)}(a)}{4!}(x-a)^4$$

$$= 2^4 + 4\cdot 2^3(x-2) + 12\cdot \frac{2^2}{2}(x-2)^2 + 24\cdot \frac{2}{6}(x-2)^3 + \frac{24}{24}(x-2)^4$$

$= 16 + 32(x-2) + 24(x-2)^2 + 8(x-2)^3 + (x-2)^4$. It is a finite series, so it converges everywhere.

10. First we need the series expansion for $\sin(x^2)$, which is

$$x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} \mp\ldots$$

Now plug this into the fraction in the problem to get

$$\frac{x^2 - \frac{x^6}{6} + \frac{x^{10}}{5!} \mp \dots - x^2}{x^6} = \frac{1}{6} + \frac{x^4}{5!} +\ldots \to \frac{1}{6},$$

that is, the limit is 1/6.

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