Difference between revisions of "1950 AHSME Problems/Problem 39"

(soln)
(Problem)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
Given the series <math> 2\plus{}1\plus{}\frac {1}{2}\plus{}\frac {1}{4}\plus{}...</math> and the following five statements:
+
Given the series <math> 2+1+\frac {1}{2}+\frac {1}{4}+\cdots</math> and the following five statements:
 
* (1) the sum increases without limit
 
* (1) the sum increases without limit
 
* (2) the sum decreases without limit
 
* (2) the sum decreases without limit
Line 23: Line 23:
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 15:13, 9 May 2015

Problem

Given the series $2+1+\frac {1}{2}+\frac {1}{4}+\cdots$ and the following five statements:

  • (1) the sum increases without limit
  • (2) the sum decreases without limit
  • (3) the difference between any term of the sequence and zero can be made less than any positive quantity no matter how small
  • (4) the difference between the sum and 4 can be made less than any positive quantity no matter how small
  • (5) the sum approaches a limit

Of these statments, the correct ones are:

$\textbf{(A)}\ \text{Only }3 \text{ and }4\qquad \textbf{(B)}\ \text{Only }5 \qquad \textbf{(C)}\ \text{Only }2\text{ and }4 \qquad \textbf{(D)}\ \text{Only }2,3\text{ and }4 \qquad \textbf{(E)}\ \text{Only }4\text{ and }5$

Solution

This series is a geometric series with common ratio $\frac{1}{2}$. Using the well-known formula for the sum of an infinite geometric series, we obtain that this series has a value of $2\cdot \frac{1}{1-\frac{1}{2}}=4$. It immediately follows that statements 1 and 2 are false while statements 4 and 5 are true. 3 is false because we cannot make any term of the sequence approach 0, though we can choose some term that is less than any given positive quantity. The correct answer is therefore $\boxed{\textbf{(E)}\ \text{Only }4\text{ and }5}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 38
Followed by
Problem 40
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png