## anonymous 3 years ago show that the sequence xn=1/n+1+1/n+2+---+1/n+n converge

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1. anonymous

pls solve this problem

2. perl

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3. anonymous

yes that is the problem

4. 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:$\int\limits_{0}^1\frac{1}{1+x}dx$

5. anonymous

thank you next question

6. anonymous

expand xy2+cos[xy] upto fourth degree terms using maclaurin's expansion

7. perl

@joemath314159 hello, how didyou get that answer

8. perl

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9. perl

thats as far as i got

10. anonymous

11. perl

joe got it, i was wondering how

12. perl

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13. perl

not only does it converge but we can find what it converges to

14. anonymous

ln2 is the answer or what

15. perl

it only asked if it converged . and yes it does

16. anonymous

joe is correct or yours

17. 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.

18. anonymous

19. perl

joemath, i see that. im just not sure how you came to this solution.

20. perl

i will check it, thanks

21. perl

and is there another way to solve this?

22. 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.

23. perl

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24. anonymous

Thats right. Let a = 0 and b = 1.

25. perl

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26. 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?

27. perl

expand xy2+cos[xy] upto fourth degree terms using maclaurin's expansion

28. anonymous

ya thats the second question

29. perl

we know that cos x = 1 - x^2/2!+ x^4/4!+...

30. anonymous

31. perl

cya :)

32. perl

@nethravathy |dw:1366722117226:dw|

33. anonymous

ya the same

34. perl

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