Consider the following series. You are expected to do all of your work on scrap
ID: 2842332 • Letter: C
Question
Consider the following series. You are expected to do all of your work on scrap paper and once you have come up with your answer you should usecopies of the accompanying answer sheet to do each of the following: If a series is a geometric series find the sum of the series. If a series is a telescoping series, find the expression sn and the sum of the series. Otherwise Frove if they are absolutely convergent, conditionally convergent or divergent. State your conclusion in an answer box. State in the answer box what test you used. If you use the integral test, you must show that you have verified that the function f(x) is decreasing. If you use the comparison test or the limit comparison test, clearly state what series you are comparing it to. You must use copies of page 3 when writing up your final answer to the problems in part II. If you misuse the divergence test and say that since the converges, you will receive no credit. Conlusion answer box: Give the letter choice of the correct answer. The series is: Absolutely Convergent Conditionally Conevergent Divert List the test you used to determine whether or not the series is Absolut snce If not absolutely convergent, list the test you used to determine if the series is Conditionally Convergence: Is the series geometric of telescoping (yes or no) ? If the series geometric or telescoping, what is its sum?Explanation / Answer
a(n) = sin(n)/n^2
|a(n)| = |sin(n)/n^2| = |sin(n)| / n^2
Note that since n^2 is always positive, |n^2| = n^2
We know that sine has a range of -1 to 1: -1 <= sin(n) <= 1
If we take the absolute value, sine can only take on values from 0 to 1: 0 <= |sin(n)| <= 1
Dividing by n^2 gives the inequality: 0/n^2 <= |sin(n)|/n^2 <= 1/n^2
Or 0 <= |sin(n)|/n^2 <= 1/n^2
We are testing |a(n)| so we will be testing for absolute convergence. If |a(n)| converges, then a(n) will be absolutely convergent.
Since we were able to set up an inequality, direct comparison test is a good idea. We will compare |a(n)| = |sin(n)|/n^2 to b(n) = 1/n^2, a convergent p-series.
Since the comparison series b(n) = 1/n^2 is convergent and |a(n)| <= b(n) is true, |a(n)| will converge by the direct comparison test. Since |a(n)| converges, a(n) = sin(n)/n^2 is absolutely convergent.
The series is neither geometric (of ar^(n-1) form) or telescoping (can be rewritten with partial fractions to cancel out terms).