anonymous
  • anonymous
show that the sequence xn=1/n+1+1/n+2+---+1/n+n converge
Linear Algebra
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anonymous
  • anonymous
show that the sequence xn=1/n+1+1/n+2+---+1/n+n converge
Linear Algebra
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
pls solve this problem
perl
  • perl
|dw:1366719299230:dw|
anonymous
  • anonymous
yes that is the problem

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anonymous
  • anonymous
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:\[\int\limits_{0}^1\frac{1}{1+x}dx\]
anonymous
  • anonymous
thank you next question
anonymous
  • anonymous
expand xy2+cos[xy] upto fourth degree terms using maclaurin's expansion
perl
  • perl
@joemath314159 hello, how didyou get that answer
perl
  • perl
|dw:1366720645769:dw|
perl
  • perl
thats as far as i got
anonymous
  • anonymous
is this the answer
perl
  • perl
joe got it, i was wondering how
perl
  • perl
|dw:1366720894172:dw|
perl
  • perl
not only does it converge but we can find what it converges to
anonymous
  • anonymous
ln2 is the answer or what
perl
  • perl
it only asked if it converged . and yes it does
anonymous
  • anonymous
joe is correct or yours
anonymous
  • anonymous
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:\[\sum_{k=1}^{n}f\left(\frac{k}{n}\right)\cdot\left(\frac{k}{n}-\frac{k-1}{n}\right)\].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.
anonymous
  • anonymous
Skip to 4:30 of this video, it should explain the idea with a similar problem. http://www.youtube.com/watch?v=xt86sFvvdhA
perl
  • perl
joemath, i see that. im just not sure how you came to this solution.
perl
  • perl
i will check it, thanks
perl
  • perl
and is there another way to solve this?
anonymous
  • anonymous
im not sure if there is another way to get the actual limit. This might be the only way to get it by hand.
perl
  • perl
|dw:1366721812063:dw|
anonymous
  • anonymous
Thats right. Let a = 0 and b = 1.
perl
  • perl
|dw:1366721876414:dw|
perl
  • perl
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?
perl
  • perl
expand xy2+cos[xy] upto fourth degree terms using maclaurin's expansion
anonymous
  • anonymous
ya thats the second question
perl
  • perl
we know that cos x = 1 - x^2/2!+ x^4/4!+...
anonymous
  • anonymous
I haven't been thinking about it =/ I'm getting ready to head to class >.<
perl
  • perl
cya :)
perl
  • perl
@nethravathy |dw:1366722117226:dw|
anonymous
  • anonymous
ya the same
perl
  • perl
|dw:1366722262487:dw|

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