Determining the limit of a sequence

  • #1
I have the following sequence

[tex]

\begin{array}{l}
a_n = ( - 1)^n \left( {\frac{n}{{n + 1}}} \right) \\
\\
\mathop {\lim }\limits_{n \to \infty } \left[ {( - 1)^n \left( {\frac{n}{{n + 1}}} \right)} \right] \\
\end{array}



[/tex]

Direct substitution yields [tex]
( - 1)^\infty \left( {\frac{\infty }{\infty }} \right)
[/tex]

I tried manipulating it into a form in which I could apply L'Hopital's Rule.

[tex]


\displaylines{
{\rm Let y} = \mathop {\lim }\limits_{n \to \infty } \left[ {( - 1)^n \left( {\frac{n}{{n + 1}}} \right)} \right] \cr
\cr
\ln y = \mathop {\lim }\limits_{n \to \infty } \ln \left[ {( - 1)^n \left( {\frac{n}{{n + 1}}} \right)} \right] \cr
\cr
= \mathop {\lim }\limits_{n \to \infty } \left[ {\ln ( - 1)^n + \ln (n) - \ln (n + 1)} \right] \cr
\cr
= \mathop {\lim }\limits_{n \to \infty } \left[ {n\ln ( - 1) + \ln (n) - \ln (n + 1)} \right] \cr
\cr
\ln ( - 1) = undefined \cr}


[/tex]

The answer is below. How did the book arrive at that answer? How did they go through and calculate the limit? Solutions manuals are so wonderfully detailed :)

http://img70.imageshack.us/img70/7812/answer5ck.jpg [Broken]
 
Last edited by a moderator:

Answers and Replies

  • #2
308
0
[tex](\frac{n}{n+1})=(1-\frac{1}{n+1})[/tex]
hope that helps
 
  • #3
335
4
opticaltempest said:
I have the following sequence

[tex]

\begin{array}{l}
a_n = ( - 1)^n \left( {\frac{n}{{n + 1}}} \right) \\
\\
\mathop {\lim }\limits_{n \to \infty } \left[ {( - 1)^n \left( {\frac{n}{{n + 1}}} \right)} \right] \\
\end{array}



[/tex]

Direct substitution yields [tex]
( - 1)^\infty \left( {\frac{\infty }{\infty }} \right)
[/tex]

I tried manipulating it into a form in which I could apply L'Hopital's Rule.

[tex]


\displaylines{
{\rm Let y} = \mathop {\lim }\limits_{n \to \infty } \left[ {( - 1)^n \left( {\frac{n}{{n + 1}}} \right)} \right] \cr
\cr
\ln y = \mathop {\lim }\limits_{n \to \infty } \ln \left[ {( - 1)^n \left( {\frac{n}{{n + 1}}} \right)} \right] \cr
\cr
= \mathop {\lim }\limits_{n \to \infty } \left[ {\ln ( - 1)^n + \ln (n) - \ln (n + 1)} \right] \cr
\cr
= \mathop {\lim }\limits_{n \to \infty } \left[ {n\ln ( - 1) + \ln (n) - \ln (n + 1)} \right] \cr
\cr
\ln ( - 1) = undefined \cr}


[/tex]

The answer is below. How did the book arrive at that answer? How did they go through and calculate the limit? Solutions manuals are so wonderfully detailed :)

http://img70.imageshack.us/img70/7812/answer5ck.jpg [Broken]
[/URL]

Are you required to test the series this way? I would suggest looking at [tex] \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} [/tex] in which case that nasty power of n cancels out. If the limit above approaches -1, then the series limit is undetermined. (It may converge, it may not.)

-Dan
 
Last edited by a moderator:
  • #4
benorin
Homework Helper
Insights Author
Gold Member
1,281
86
It is a sequence opposed to a series that is being tested for convergence, and hence only [tex]\lim_{n\rightarrow\infty}a_n[/tex] need be considered (and for such happyg1's note ought be sufficient.)
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
956
Another way to look at it is to divide both numerator and denominator by n:
[tex](-1)^n\frac{n}{n+1}= (-1)^n\frac{1}{1+\frac{1}{n}}[/tex]

As n-> infinity, 1/n-> 0.
 

Related Threads on Determining the limit of a sequence

Replies
5
Views
2K
Replies
14
Views
2K
Replies
1
Views
601
  • Last Post
Replies
2
Views
709
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
9
Views
620
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
1
Views
859
Top