Determining the limit of a sequence

[opticaltempest]

I have the following sequence

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

Direct substitution yields <br /> ( - 1)^\infty \left( {\frac{\infty }{\infty }} \right)<br />

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

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

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

 

[happyg1]

(\frac{n}{n+1})=(1-\frac{1}{n+1})
hope that helps

 

[topsquark]

I have the following sequence

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

Direct substitution yields <br /> ( - 1)^\infty \left( {\frac{\infty }{\infty }} \right)<br />

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

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

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

[/URL]

Are you required to test the series this way? I would suggest looking at \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} 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

 

[benorin]

It is a sequence opposed to a series that is being tested for convergence, and hence only \lim_{n\rightarrow\infty}a_n need be considered (and for such happyg1's note ought be sufficient.)

 

[HallsofIvy]

Another way to look at it is to divide both numerator and denominator by n:
(-1)^n\frac{n}{n+1}= (-1)^n\frac{1}{1+\frac{1}{n}}

As n-> infinity, 1/n-> 0.