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nethravathy
Group Title
show that the sequence xn=1/n+1+1/n+2++1/n+n converge
 one year ago
 one year ago
nethravathy Group Title
show that the sequence xn=1/n+1+1/n+2++1/n+n converge
 one year ago
 one year ago

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nethravathy Group TitleBest ResponseYou've already chosen the best response.0
pls solve this problem
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
dw:1366719299230:dw
 one year ago

nethravathy Group TitleBest ResponseYou've already chosen the best response.0
yes that is the problem
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.1
You can rewrite the series as:\[\frac{\frac{1}{n}}{1+\frac{1}{n}}+\frac{\frac{1}{n}}{1+\frac{2}{n}}+\frac{\frac{1}{n}}{1+\frac{3}{n}}+\cdots+\frac{\frac{1}{n}}{1+\frac{n}{n}}\]\[=\frac{1}{1+\frac{1}{n}}\cdot \frac{1}{n}+\frac{1}{1+\frac{2}{n}}\cdot \frac{1}{n}+\cdots +\frac{1}{1+\frac{n}{n}}\cdot \frac{1}{n}\]Then it should be clear that this series is actually a Riemann sum. Its getting the areas under the function:\[f(x)=\frac{1}{1+x}\] over the interval [0,1] by splitting the interval into n equal pieces. So the limit as n goes to infinity is:\[\int\limits_{0}^1\frac{1}{1+x}dx\]
 one year ago

nethravathy Group TitleBest ResponseYou've already chosen the best response.0
thank you next question
 one year ago

nethravathy Group TitleBest ResponseYou've already chosen the best response.0
expand xy2+cos[xy] upto fourth degree terms using maclaurin's expansion
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
@joemath314159 hello, how didyou get that answer
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
dw:1366720645769:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
thats as far as i got
 one year ago

nethravathy Group TitleBest ResponseYou've already chosen the best response.0
is this the answer
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
joe got it, i was wondering how
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
dw:1366720894172:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
not only does it converge but we can find what it converges to
 one year ago

nethravathy Group TitleBest ResponseYou've already chosen the best response.0
ln2 is the answer or what
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
it only asked if it converged . and yes it does
 one year ago

nethravathy Group TitleBest ResponseYou've already chosen the best response.0
joe is correct or yours
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.1
You want to look at it as:\[x_n=\sum_{k=1}^{n}\frac{1}{n+k}=\sum_{k=1}^{n}\frac{\frac{1}{n}}{1+\frac{k}{n}}=\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}\cdot\frac{1}{n}\]Then let:\[f(x)=\frac{1}{1+x}\]Using this, we can change the sum to:\[\sum_{k=1}^{n}f\left(\frac{k}{n}\right)\cdot\left(\frac{k}{n}\frac{k1}{n}\right)\].This is a Riemann sum of f(x) over the interval [0,1]. If you split the interval into n equal pieces, each piece is 1/n long. so thats the width of a rectangle. The height is the value of the functions at the point x = k/n.
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.1
Skip to 4:30 of this video, it should explain the idea with a similar problem. http://www.youtube.com/watch?v=xt86sFvvdhA
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
joemath, i see that. im just not sure how you came to this solution.
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
i will check it, thanks
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
and is there another way to solve this?
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.1
im not sure if there is another way to get the actual limit. This might be the only way to get it by hand.
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
dw:1366721812063:dw
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.1
Thats right. Let a = 0 and b = 1.
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
dw:1366721876414:dw
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
well you are a genius to have solved this. this is not obvious to use this approach. any luck with the second question the maclaurin?
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
expand xy2+cos[xy] upto fourth degree terms using maclaurin's expansion
 one year ago

nethravathy Group TitleBest ResponseYou've already chosen the best response.0
ya thats the second question
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
we know that cos x = 1  x^2/2!+ x^4/4!+...
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.1
I haven't been thinking about it =/ I'm getting ready to head to class >.<
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
@nethravathy dw:1366722117226:dw
 one year ago

nethravathy Group TitleBest ResponseYou've already chosen the best response.0
ya the same
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
dw:1366722262487:dw
 one year ago
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