## LogicalApple 2 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.

1. LogicalApple

2. LogicalApple

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

3. duplicitycheese

@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

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

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

6. joemath314159

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

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

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

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

10. LogicalApple

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

11. LogicalApple

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

12. nitz

correct.......

13. LogicalApple

Whew, thank you!

14. nitz

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

15. nitz

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

16. LogicalApple

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

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

18. nitz

actually we had some other way of doing these questions