# Determining the limit of a sequence

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 :)

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## Answers and Replies

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

opticaltempest said:
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 :)

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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

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benorin
Homework Helper
Gold Member
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
$$(-1)^n\frac{n}{n+1}= (-1)^n\frac{1}{1+\frac{1}{n}}$$