Question

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.

Answer 1

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

Answer 2

@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?

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

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!

Answer 7

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

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.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

correct.......

Answer 12

Whew, thank you!

Answer 13

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

Answer 14

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

Answer 15

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

\[f(x _{i})\]
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.

Answer 16

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

Answer 17

actually we had some other way of doing these questions