Determining the limit of a sequence

1. Feb 20, 2006

opticaltempest

I have the following sequence

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

Direct substitution yields $$( - 1)^\infty \left( {\frac{\infty }{\infty }} \right)$$

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

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

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 :)

Last edited by a moderator: May 2, 2017
2. Feb 20, 2006

happyg1

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

3. Feb 20, 2006

topsquark

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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

Last edited by a moderator: May 2, 2017
4. Feb 20, 2006

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.)

5. Feb 21, 2006

HallsofIvy

Staff Emeritus
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.