Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Linear Algebra
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

pls solve this problem
|dw:1366719299230:dw|
yes that is the problem

Not the answer you are looking for?

Search for more explanations.

Ask your own question

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
|dw:1366720645769:dw|
thats as far as i got
is this the answer
joe got it, i was wondering how
|dw:1366720894172:dw|
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. http://www.youtube.com/watch?v=xt86sFvvdhA
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.
|dw:1366721812063:dw|
Thats right. Let a = 0 and b = 1.
|dw:1366721876414:dw|
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
|dw:1366722262487:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question