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

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

joemath314159
 one year ago
Best ResponseYou've already chosen the best response.1You 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:\[\int\limits_{0}^1\frac{1}{1+x}dx\]

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

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

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

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

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

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

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

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

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

joemath314159
 one year ago
Best ResponseYou've already chosen the best response.1You 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.

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

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

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

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

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

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

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

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

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

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

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