calculus- find an explicit formula for a(n)
\[a _{1}=\frac{ 1 }{ 2-\frac{ 1 }{ 2 }} , a _{2}=\frac{ 2 }{ 3-\frac{ 1 }{ 3}} , a _{3}=\frac{ 3 }{ 4-\frac{ 1 }{ 4}}....a _{5}\]
find lim as n approaches infinity of a(n)
is it not clear ?
what would \(a_4\) be? if you know that, then you know them all
\[a _{4}= \frac{ 4 }{ 5-\frac{ 1 }{ 5 } }\]
yeah ok so you got it
but how do i write a formula?
\[a_{50}=?\]
\[a _{50}=\frac{ 50 }{ 51- \frac{ 1 }{51 }}\]
sure so if you can do it with \(50\) you can certainly do it with \(n\) right?
oh btw i think you made a mistake on that one i think what you wrote was \[a_{51}\]
oh no, i made a mistake, not you
oh i got it!
i mean i can tell you the answer, but visualize \(n\) instead of \(50\)
what goes up, what goes down?
n/((n+1)-(1/(n+1)))
lol yeah that one
\[\frac{n}{(n+1)+\frac{1}{n+1}}\]
how do i find lim as n approaches infinity of a(n)
limit should be obvious
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
as \(n\to\infty \) \(\frac{1}{n+1}\to 0\) right ?
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
by which i mean, get rid of the compound fraction by multiplying top and bottom by \(n+1\) to get a rational expression
thank you
yw
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