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Mathematics 16 Online
OpenStudy (darkigloo):

calculus- find an explicit formula for a(n)

OpenStudy (darkigloo):

\[a _{1}=\frac{ 1 }{ 2-\frac{ 1 }{ 2 }} , a _{2}=\frac{ 2 }{ 3-\frac{ 1 }{ 3}} , a _{3}=\frac{ 3 }{ 4-\frac{ 1 }{ 4}}....a _{5}\]

OpenStudy (darkigloo):

find lim as n approaches infinity of a(n)

satellite73 (satellite73):

is it not clear ?

satellite73 (satellite73):

what would \(a_4\) be? if you know that, then you know them all

OpenStudy (darkigloo):

\[a _{4}= \frac{ 4 }{ 5-\frac{ 1 }{ 5 } }\]

satellite73 (satellite73):

yeah ok so you got it

OpenStudy (darkigloo):

but how do i write a formula?

satellite73 (satellite73):

\[a_{50}=?\]

OpenStudy (darkigloo):

\[a _{50}=\frac{ 50 }{ 51- \frac{ 1 }{51 }}\]

satellite73 (satellite73):

sure so if you can do it with \(50\) you can certainly do it with \(n\) right?

satellite73 (satellite73):

oh btw i think you made a mistake on that one i think what you wrote was \[a_{51}\]

satellite73 (satellite73):

oh no, i made a mistake, not you

OpenStudy (darkigloo):

oh i got it!

satellite73 (satellite73):

i mean i can tell you the answer, but visualize \(n\) instead of \(50\)

satellite73 (satellite73):

what goes up, what goes down?

OpenStudy (darkigloo):

n/((n+1)-(1/(n+1)))

satellite73 (satellite73):

lol yeah that one

satellite73 (satellite73):

\[\frac{n}{(n+1)+\frac{1}{n+1}}\]

OpenStudy (darkigloo):

how do i find lim as n approaches infinity of a(n)

satellite73 (satellite73):

limit should be obvious

satellite73 (satellite73):

but i see that it is not, so lets do some number therapy first, which for some reason most people don't like i put \(n=100\) which is not that big, check this http://www.wolframalpha.com/input/?i=%28100%29%2F%28101%2B1%2F101%29

satellite73 (satellite73):

as \(n\to\infty \) \(\frac{1}{n+1}\to 0\) right ?

satellite73 (satellite73):

so you are left with \[\frac{n}{n+1}\] and that limit is clear if you want to spend a bunch of time, you could do the algebra to see what you get

satellite73 (satellite73):

by which i mean, get rid of the compound fraction by multiplying top and bottom by \(n+1\) to get a rational expression

OpenStudy (darkigloo):

thank you

satellite73 (satellite73):

yw

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