The Student Room Group

Limit of a sequence

{xn} is a sequence defined by xn=(−7)nn3+12⋅8n+n48n+2+n6x_{n}=\frac{(-7)^nn^3+12\cdot 8^n+n^4}{8^{n+2}+n^6}. I need to show {xn} converges and find its limit.

I tried to follow a similar example and divided the top and bottom of the fraction by 8nn68^nn^6, and I obtained:
xn=(−78)nn−3+12n−6+n−282n−6+(18)nx_{n}=\frac{(\frac{-7}{8})^nn^{-3}+12n^{-6}+n^{-2}}{8^{2}n^{-6}+(\frac{1}{8})^n}

But when I applied some standard results (e.g. (-7/8)^n -> 0), I ended up with division by 0.:frown:

Any hints would be appreciated!
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Avatar for Blazy
Blazy
8
Original post by rayquaza17
{xn} is a sequence defined by xn=(−7)nn3+12⋅8n+n48n+2+n6x_{n}=\frac{(-7)^nn^3+12\cdot 8^n+n^4}{8^{n+2}+n^6}. I need to show {xn} converges and find its limit.

I tried to follow a similar example and divided the top and bottom of the fraction by 8nn68^nn^6, and I obtained:
xn=(−78)nn−3+12n−6+n−282n−6+(18)nx_{n}=\frac{(\frac{-7}{8})^nn^{-3}+12n^{-6}+n^{-2}}{8^{2}n^{-6}+(\frac{1}{8})^n}

But when I applied some standard results (e.g. (-7/8)^n -> 0), I ended up with division by 0.:frown:

Any hints would be appreciated!


Try dividing by 8n 8^n instead. You may need to should justify why (−78)nn3 (-\frac{7}{8})^n n^3 , nk8n \frac{n^k}{8^n} tends to 0.
(edited 9 years ago)
Reply 2
Avatar for rayquaza17
rayquaza17
OP
17
Thank you. :smile:

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