A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Riemann Sum . . .
Attached is the problem of evaluating a series as n approaches infinity.
I am stuck on how to continue. I know if I can sqrt(n / (n + 1)) in a different form where there is only one n, I can do the rest.
anonymous
 4 years ago
Riemann Sum . . . Attached is the problem of evaluating a series as n approaches infinity. I am stuck on how to continue. I know if I can sqrt(n / (n + 1)) in a different form where there is only one n, I can do the rest.

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That is to say, if I can get sqrt(n / (n + k)) to a form with only one n in it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@logicalApple, I'm not sure either with the rest of the expression but if you have 1/n at the front and n is approaching infinite, can the whole thing just be approaching zero?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01/n would approach 0 but that wouldn't cause the sum to approach 0. What if what is inside the parentheses approaches infinity? Then we have an indeterminate form 0 * infinity. If we can frame this into a Riemann Sum then it becomes a (hopefully) simple integration problem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hm, I didn't think of that. I have no idea then. Sorry!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Maybe change the expression to:\[\sqrt{\frac{n}{n+k}}=\sqrt{\frac{(n+k)k}{n+k}}=\sqrt{\frac{n+k}{n+k}\frac{k}{n+k}}\]\[=\sqrt{1\frac{k}{n+k}}\]This expresion has only one n in it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hm... now there are two k's. I wonder if it's possible to write it in a form where there is only one of each? Otherwise I am not sure how to continue. I will work on this later. Thanks for the suggestions!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ n }{ n+k } = \frac{ 1 }{ a + b } > \frac{ n+k }{ n } = a + b > 1 + \frac{ k }{ n } = a + b\] So . . . \[\frac{ n }{ n+k } = \frac{ 1 }{ 1 + \frac{ k }{ n } }\] Now the series becomes more manageable as: \[\lim_{n \rightarrow \infty} \frac{ 1 }{ n } \left( \frac{ 1 }{ \sqrt{1 + \frac{ 1 }{ n }} } + \frac{ 1 }{ \sqrt{1 + \frac{ 2 }{ n }} } + \frac{ 1 }{ \sqrt{1 + \frac{ 3 }{ n }} } + ... + \frac{ 1 }{ \sqrt{1 + \frac{ n }{ n }} } \right)\] Now I can imagine a function, f(x) = 1/sqrt(1 + x) divided into n partitions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} \frac{ 1 }{ n }\sum_{1}^{n}\left( f(\frac{ i }{ n }) \right), f(x) = \frac{ 1 }{ \sqrt{1+ x} }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1}\frac{ 1 }{ \sqrt{1+x} }dx = 2(\sqrt{2}1)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm pretty sure of it. I wonder if someone can verify. Wolframalpha does not evaluate infinite series as well as humans.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0till generalising it was fine...........but actually didnt get your last step.....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how can u integrate.......?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} \frac{ 1 }{ n }\sum_{i =1}^{n}\left( f(\frac{ i }{ n }) \right), f(x) = \frac{ 1 }{ \sqrt{1+ x} }\] \[f(x _{i})\] returns the ith element of the sum \[\frac{ 1 }{ n } = \Delta x\] This is our partition with b  a = 1, where a is the lower limit and b is the upper limit. Let a = 0 and b = 1, then 0 < 1/n < 2/n < ... < i/n < ... < (n1)/n < b \[a = 0 < x _{1} = \frac{ 1 }{ n } < x _{2} = \frac{ 2 }{ n } < ... < x _{i} = \frac{ i }{ n }, ... < x _{n1} = \frac{ n1 }{ n } < 1 = b\] So all these points of the function are partitioned by 1/n. I think this represents a Riemann sum as \[\sum_{i = 1}^{n} f(x _{i}) \Delta x\] Which, as n approaches infinity, becomes the Reimann integral: \[\int\limits_{0}^{1} f(x) dx\] where \[f(x) = \frac{ 1 }{ \sqrt{1 + x} }\] Sorry if my explanation is bad. I was never good with summations.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok.............right ........thanks

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually we had some other way of doing these questions
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.