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    We just started the limits of sequences and series today, and i am so confused. How would I go about finding the limit?

    \displaystyle\lim_{n\to{\infty}} (1+ \frac{-1^n}{n})

    *edit*

    I know the therom that states

    If |r| < 1. then \displaystyle\lim_{n\to{\infty}}  r^n = 0
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    limit = lim 1 + lim [-1^n]/n

    lim [-1^n]/n

    Think about what happens as n gets large... It would oscillate, but think about what happens with a large n at the bottom...

    lim [-1^n]/n what does that tend to as n --- > infinity]

    EDIT: The theorem doesn't help you |r| </ 1
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    i still dont get it.
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    Think of  \displaystyle \lim_ {n \to {\infty}} \frac{1}{n}
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    ok... do you know how to find the limit of [-1^n]/n?

    -1^LARGE NUMBER = either -1 or +1 depending if the large number is even or odd

    but -1/LARGE NUMBER and +1/LARGE NUMBER still lead to the same conclusion...
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    so its a limit of 0?
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    Not for the entire thing, no.

    limit = 1 + lim [-1^n]/n
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    Yeah just that part, now just add the limts
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    (Original post by nk9230)
    We just started the limits of sequences and series today, and i am so confused. How would I go about finding the limit?

    \displaystyle\lim_{n\to{\infty}} (1+ \frac{-1^n}{n})

    *edit*

    I know the therom that states

    If |r| &lt; 1. then \displaystyle\lim_{n\to{\infty}}  r^n = 0
    If you want a way that shows the limit of  \frac{-1^n}{n} then you could take the absolute value, or you could square it and use the product rule, i.e. \displaystyle\lim_{n\to{\infty}}  \frac{1^2^n}{n^2} which is easier to see that it tends to 0. Then by the product rule and uniqueness of limits  \frac{-1^n}{n} must tend to 0, and you can use the sum rule to find the limit of the whole thing.
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    (-1)^n is either -1 or 1

    [(-1)^n]/n is either -1/n or 1/n

    When n is absolutely huge, -1/n and 1/n are both pretty close to zero.

    So: lim 1 + [(-1)^n]/n

    = 1 + lim [(-1)^n]/n

    = 1 + 0

    1 + 0 = ... I'll let you work that one out :p:
 
 
 
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