CARNEGIE MELLON UNIVERSITY
DEPARTMENT OF MATHEMATICAL SCIENCES
21-256 Solutions, Review Exam 1, Spring 2004

1. Determine if the following sequence converges or diverges. If it converges find the limit.

   
   

   $$a_n= \frac{2n^2+e^{-n}}{n^2+\cos n}$$
   
   
   Ans: Divide through by $n^2$. The term $\cos n/n^2$ goes to $0$ by the squeeze theorem.
   $$\begin{eqnarray*} \lim_{n\rightarrow \infty} a_n & = & \lim_{n\rightarrow \in... ...}{1+\frac{\cos n}{n^2}} \\ & = & \frac{2+0}{1+0} \\ & = & 2 \end{eqnarray*}$$
   
   
   
   
   

2. Determine if the following series converges or diverges. If it converges find the sum.
   
   $$\sum_{n=1}^{\infty} (-4)^{1-n} (\pi)^{n}$$
   
   
   Ans: The series is geometric with $a=\pi$, and $r=-\pi/4$. Since $\vert r\vert <1$, the series converges to
   $$\begin{eqnarray*} \frac{a}{1-r} & = & \frac{\pi}{1+\pi/4} \\ & = & \frac{4\pi}{4+\pi } \end{eqnarray*}$$
   
   
   
   
   

3. Determine if the following series converges or diverges.
   
   $$\sum_{n=1}^{\infty} \frac {\sqrt[4] {n^3+n^2+1} }{\sqrt[5]{1-n+n^9}}$$
   
   
   Ans: We apply the limit comparison test, using $b_n=\frac{n^{3/4}}{n^{9/5}}$
   $$\begin{eqnarray*} \lim_{n\rightarrow \infty} \frac{a_n}{b_n} & = & \lim_{n\ri... ...1/4} \left[ \frac{1}{1/n^9-1/n^8+1} \right]^{1/5} \\ & =& 1 \end{eqnarray*}$$
   
   
   
   
   

   Since $b_n=\frac{1}{n^{9/5-3/4}}=\frac{1}{n^{21/20}}$ determines a convergent p-series, our series converges.

4. Determine if the following series converges conditionally, converges absolutely, or diverges.
   
   $$\sum_{n=2}^{\infty} (-1)^{n-1} \frac{1}{n \ln n}$$
   
   
   Ans: The series converges conditionally. The Alternating Series Test, with $b_n=1/(n \ln n)$ is used to show that the series converges. The Integral Test is used to show that the series does not converge absolutely.

5. Determine if the following series converges conditionally, converges absolutely, or diverges.
   
   $$\sum_{n=1}^{\infty} \frac{\sin^n n}{n^n}$$
   
   
   Ans: The series converges absolutely using the Root Test. The $n^{\rm th}$ root of the absolute value is $\frac{\vert\sin n\vert}{n}$, which converges to $0<1$ using the squeeze theorem.

6. Determine if the following series converges conditionally, converges absolutely, or diverges.
   
   $$\sum_{n=0}^{\infty } \frac{(-e)^n}{n!}$$
   
   

   Ans: The series converges absolutely using the Ratio Test. The ratio $\frac{\vert a_{n+1}\vert}{\vert a_n\vert}$ simplifies to $\frac{e}{n+1}$, which converges to $0<1$. Also, using the Taylor Series for $e^x$, we know the sum of this series is $e^{-e}$.

7. Find the interval of convergence of the following series - you do not need to test endpoints.
   
   $$\sum_{n=1}^{\infty } n^2 4^n (x+3)^{2n}$$
   
   
   Ans: Apply the Ratio Test:
   $$\begin{eqnarray*} \frac{\vert a_{n+1}\vert}{\vert a_n\vert} & = & \frac{(n+1)... ... \\ & = & 4 \left[ \frac{n+1}{n} \right]^2 \vert x+3\vert^2, \end{eqnarray*}$$
   
   
   
   
   
   which converges to $4\vert x+3\vert^2$. Set this less that $1$ and solve for $x$, we get $\vert x+3\vert<1/2$, so $-7/2 <x < -5/2$.

8. Find a power series representation and interval of convergence of
   
   $$f(x)=\frac{ 2x^2}{4x+3} .$$
   
   
   Ans: Write $f(x)=\frac{2x^2}{3}\frac{1}{1-(-4x/3)}$. Then
   $$\begin{eqnarray*} f(x) & = & \frac{2x^2}{3} \sum_{n=0}^{\infty} (-4x/3)^n \\ ... ... & = & \sum_{n=0}^{\infty} 2\frac{(-1)^n4^nx^{n+2}}{3^{n+1}} \end{eqnarray*}$$
   
   
   
   
   

   The interval of convergence is obtained by setting $\vert-4x/3\vert<1$, we get $-3/4<x<3/4$.

9. Find the Taylor series centered at $a=0$ for $f(x)=xe^{x^2}$. You may use the known series for $e^x$.

   Ans: Using the known series for $e^x$, we get

   $$\begin{eqnarray*} f(x) & = & x \sum_{n=0}^{\infty} \frac{ (x^2)^n}{n!} \\ & =... ...{2n} }{n!} \\ & = & \sum_{n=0}^{\infty} \frac{ x^{2n+1} }{n!} \end{eqnarray*}$$