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LogicalApple
Group Title
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.
 one year ago
 one year ago
LogicalApple Group Title
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.
 one year ago
 one year ago

This Question is Closed

LogicalApple Group TitleBest ResponseYou've already chosen the best response.2
That is to say, if I can get sqrt(n / (n + k)) to a form with only one n in it.
 one year ago

duplicitycheese Group TitleBest 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?
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.2
1/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.
 one year ago

duplicitycheese Group TitleBest ResponseYou've already chosen the best response.0
hm, I didn't think of that. I have no idea then. Sorry!
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.1
Maybe 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.
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.2
Hm... 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!
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.2
\[\frac{ n }{ n+k } = \frac{ 1 }{ a + b } > \frac{ n+k }{ n } = a + b > 1 + \frac{ k }{ n } = a + b\] 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.
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.2
\[\lim_{n \rightarrow \infty} \frac{ 1 }{ n }\sum_{1}^{n}\left( f(\frac{ i }{ n }) \right), f(x) = \frac{ 1 }{ \sqrt{1+ x} }\]
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.2
\[\int\limits_{0}^{1}\frac{ 1 }{ \sqrt{1+x} }dx = 2(\sqrt{2}1)\]
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.2
I'm pretty sure of it. I wonder if someone can verify. Wolframalpha does not evaluate infinite series as well as humans.
 one year ago

nitz Group TitleBest ResponseYou've already chosen the best response.0
correct.......
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.2
Whew, thank you!
 one year ago

nitz Group TitleBest ResponseYou've already chosen the best response.0
till generalising it was fine...........but actually didnt get your last step.....
 one year ago

nitz Group TitleBest ResponseYou've already chosen the best response.0
how can u integrate.......?
 one year ago

LogicalApple Group TitleBest ResponseYou've already chosen the best response.2
\[\lim_{n \rightarrow \infty} \frac{ 1 }{ n }\sum_{i =1}^{n}\left( f(\frac{ i }{ n }) \right), f(x) = \frac{ 1 }{ \sqrt{1+ x} }\] \[f(x _{i})\] 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.
 one year ago

nitz Group TitleBest ResponseYou've already chosen the best response.0
ok.............right ........thanks
 one year ago

nitz Group TitleBest ResponseYou've already chosen the best response.0
actually we had some other way of doing these questions
 one year ago
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