# Proving this equation -- Limit of a sum of inverse square root terms

• alsdt
In summary, Wolfram showed that if you integrate the series given in Post #4, it becomes a Riemann sum.
alsdt
Homework Statement
calculate the limit of a phrase
Relevant Equations
limit
Hi
I was working on a physics problem and it was almost solved.
Only the part that is mostly mathematical remains, and no matter how hard I tried, I could not solve it.
I hope you can help me.
This is the equation I came up with and I wanted to prove it: $$\lim_{n \rightarrow+ \infty} { \sum_{i=1}^n \frac 1 {\sqrt{i(2n-i-1)} +\sqrt{i(2n-i)}}}=\frac \pi 4$$

Delta2
What have you tried?
Can you show some work?
What is the context? It might provide some insight.

I don't have a proof, but typically you have to turn these types of things into a Riemann sum. I think you can probably argue the -1 in the one term is irrelevant, at which point

https://www.wolframalpha.com/input/?i=int_0^1+1/sqrt(x(1-x))+dx

Is pretty suggestive

PeroK and mfb
Wolfram shows

So you seem to be on right track.

PeroK and robphy
As shown in post #4 you may transform the sum as
$$\sum_x \frac{\sqrt{1+x}-\sqrt{1+y}}{\sqrt{1-x}}$$
where x = ##{0,\ 1/n,\ 2/n,\ ...\ , (n-1)\ /n}## and y= x - 1/n
, which is convenient for transformation to integral.

Last edited:
anuttarasammyak said:
Wolfram shows
View attachment 291038

So you seem to be on right track.

anuttarasammyak said:
As shown in post #4 you may transform the sum as
$$\sum_x \frac{\sqrt{1+x}-\sqrt{1+y}}{\sqrt{1-x}}$$
where x = ##{0,\ 1/n,\ 2/n,\ ...\ , (n-1)\ /n}## and y= x - 1/n
, which is convenient for transformation to integral.
Thanks for editing and adding details (like what y is).
On its own, Post #4 says that Wolfram shows [something], but it wasn't clear how or what is being shown.

Are you saying that Wolfram transformed the original series into what is shown in Post #4?

robphy said:
Are you saying that Wolfram transformed the original series into what is shown in Post #4?
No, I did it by myself in a usual way.

Last edited:

## 1. What is the equation for the limit of a sum of inverse square root terms?

The equation for the limit of a sum of inverse square root terms is:
limn→∞i=1n 1/√i = ∞

## 2. How do you prove the equation for the limit of a sum of inverse square root terms?

The equation can be proven using the squeeze theorem, also known as the sandwich theorem. This involves finding two other functions that have the same limit as the original function, and using their limits to show that the original function also has the same limit.

## 3. What is the significance of the limit of a sum of inverse square root terms?

The limit of a sum of inverse square root terms is significant because it shows that the sum of infinitely many terms can still approach infinity, even if each individual term is getting smaller and smaller. This concept is important in calculus and other areas of mathematics.

## 4. Can the equation for the limit of a sum of inverse square root terms be generalized to other types of series?

Yes, the equation can be generalized to other types of series as long as they follow the same pattern of approaching infinity. For example, the limit of a sum of inverse cube root terms would also approach infinity.

## 5. Are there any real-world applications of the limit of a sum of inverse square root terms?

Yes, this concept can be applied in physics and engineering, particularly in the study of forces and fields. For example, the electric field around a point charge can be calculated using a similar equation involving the sum of inverse square root terms.

• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
17
Views
692
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
13
Views
733
• Calculus and Beyond Homework Help
Replies
3
Views
470
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
518
• Calculus and Beyond Homework Help
Replies
14
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
424
• Calculus and Beyond Homework Help
Replies
1
Views
362