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nethravathyBest ResponseYou've already chosen the best response.0
pls solve this problem
 11 months ago

nethravathyBest ResponseYou've already chosen the best response.0
yes that is the problem
 11 months ago

joemath314159Best 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:\[\int\limits_{0}^1\frac{1}{1+x}dx\]
 11 months ago

nethravathyBest ResponseYou've already chosen the best response.0
thank you next question
 11 months ago

nethravathyBest ResponseYou've already chosen the best response.0
expand xy2+cos[xy] upto fourth degree terms using maclaurin's expansion
 11 months ago

perlBest ResponseYou've already chosen the best response.0
@joemath314159 hello, how didyou get that answer
 11 months ago

nethravathyBest ResponseYou've already chosen the best response.0
is this the answer
 11 months ago

perlBest ResponseYou've already chosen the best response.0
joe got it, i was wondering how
 11 months ago

perlBest ResponseYou've already chosen the best response.0
not only does it converge but we can find what it converges to
 11 months ago

nethravathyBest ResponseYou've already chosen the best response.0
ln2 is the answer or what
 11 months ago

perlBest ResponseYou've already chosen the best response.0
it only asked if it converged . and yes it does
 11 months ago

nethravathyBest ResponseYou've already chosen the best response.0
joe is correct or yours
 11 months ago

joemath314159Best 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.
 11 months ago

joemath314159Best 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
 11 months ago

perlBest ResponseYou've already chosen the best response.0
joemath, i see that. im just not sure how you came to this solution.
 11 months ago

perlBest ResponseYou've already chosen the best response.0
and is there another way to solve this?
 11 months ago

joemath314159Best 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.
 11 months ago

joemath314159Best ResponseYou've already chosen the best response.1
Thats right. Let a = 0 and b = 1.
 11 months ago

perlBest 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?
 11 months ago

perlBest ResponseYou've already chosen the best response.0
expand xy2+cos[xy] upto fourth degree terms using maclaurin's expansion
 11 months ago

nethravathyBest ResponseYou've already chosen the best response.0
ya thats the second question
 11 months ago

perlBest ResponseYou've already chosen the best response.0
we know that cos x = 1  x^2/2!+ x^4/4!+...
 11 months ago

joemath314159Best ResponseYou've already chosen the best response.1
I haven't been thinking about it =/ I'm getting ready to head to class >.<
 11 months ago

perlBest ResponseYou've already chosen the best response.0
@nethravathy dw:1366722117226:dw
 11 months ago
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