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show that the sequence xn=1/n+1+1/n+2+---+1/n+n converge

Linear Algebra
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pls solve this problem
yes that is the problem

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Other answers:

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

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