## 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

1. LogicalApple Group Title

2. LogicalApple Group Title

That is to say, if I can get sqrt(n / (n + k)) to a form with only one n in it.

3. duplicitycheese Group Title

@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?

4. LogicalApple Group Title

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.

5. duplicitycheese Group Title

hm, I didn't think of that. I have no idea then. Sorry!

6. joemath314159 Group Title

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.

7. LogicalApple Group Title

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!

8. LogicalApple Group Title

$\frac{ n }{ n+k } = \frac{ 1 }{ a + b } --> \frac{ n+k }{ n } = a + b --> 1 + \frac{ k }{ n } = a + b$ 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.

9. LogicalApple Group Title

$\lim_{n \rightarrow \infty} \frac{ 1 }{ n }\sum_{1}^{n}\left( f(\frac{ i }{ n }) \right), f(x) = \frac{ 1 }{ \sqrt{1+ x} }$

10. LogicalApple Group Title

$\int\limits_{0}^{1}\frac{ 1 }{ \sqrt{1+x} }dx = 2(\sqrt{2}-1)$

11. LogicalApple Group Title

I'm pretty sure of it. I wonder if someone can verify. Wolframalpha does not evaluate infinite series as well as humans.

12. nitz Group Title

correct.......

13. LogicalApple Group Title

Whew, thank you!

14. nitz Group Title

till generalising it was fine...........but actually didnt get your last step.....

15. nitz Group Title

how can u integrate.......?

16. LogicalApple Group Title

$\lim_{n \rightarrow \infty} \frac{ 1 }{ n }\sum_{i =1}^{n}\left( f(\frac{ i }{ n }) \right), f(x) = \frac{ 1 }{ \sqrt{1+ x} }$ $f(x _{i})$ 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 < ... < (n-1)/n < b $a = 0 < x _{1} = \frac{ 1 }{ n } < x _{2} = \frac{ 2 }{ n } < ... < x _{i} = \frac{ i }{ n }, ... < x _{n-1} = \frac{ n-1 }{ n } < 1 = b$ 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.

17. nitz Group Title

ok.............right ........thanks

18. nitz Group Title

actually we had some other way of doing these questions