009C Sample Midterm 3

This is a department sample midterm, and is meant to represent the material usually covered in Math 9C. Click on the  boxed problem numbers  to go to a solution.

Instructions: This exam has a total of 60 points. You have 50 minutes. You must show all your work to receive full credit You may use any result done in class. The points attached to each problem are indicated beside the problem.You are not allowed books, notes, or calculators. Answers should be written as ${\sqrt {2}}$ as opposed to $1.4142135\cdots$.

Convergence and Limits of a Sequence

 Problem 1.   (12 points) Test if the following sequence ${a_{n}}$ converges or diverges. If it converges, also find the limit of the sequence.

$a_{n}=\left({\frac {n-7}{n}}\right)^{1/n}.$

Sum of a Series

 Problem 2.   For each the following series find the sum, if it converges. If you think it diverges, explain why.

(a) (6 points)      ${\frac {1}{2}}-{\frac {1}{2\cdot 3}}+{\frac {1}{2\cdot 3^{2}}}-{\frac {1}{2\cdot 3^{3}}}+{\frac {1}{2\cdot 3^{4}}}-{\frac {1}{2\cdot 3^{5}}}+\cdots .$

(b) (6 points)      ${\displaystyle \sum _{n=1}^{\infty }}{\frac {3}{(2n-1)(2n+1)}}.$

Convergence Tests for Series I

 Problem 3.   Test if each the following series converges or diverges. Give reasons and clearly state if you are using any standard test.

(a) (6 points)      ${\displaystyle \sum _{n=1}^{\infty }}\,{\frac {n!}{(3n+1)!}}.$

(b) (6 points)      ${\displaystyle \sum _{n=2}^{\infty }}\,{\frac {\sqrt {n}}{n^{2}-3}}.$

Convergence Tests for Series II

 Problem 4.   Test the series for convergence or divergence.

(a) (6 points)      ${\displaystyle \sum _{n=1}^{\infty }}\,(-1)^{n}\sin {\frac {\pi }{n}}.$

(b) (6 points)      ${\displaystyle \sum _{n=1}^{\infty }}\,(-1)^{n}\cos {\frac {\pi }{n}}$.

Radius and Interval of Convergence

 Problem 5.   Find the radius of convergence and the interval of convergence of the series.

(a) (6 points) ${\displaystyle \sum_{n=0}^{\infty}}\frac{(-1)^{n}x^{n}}{n+1}.$

(b) (6 points) ${\displaystyle \sum_{n=0}^{\infty}}\frac{(x+1)^{n}}{n^{2}}.$