# Find limit involving square of sine

• songoku
In summary: And the difference between consecutive terms of ##a_n## converges to 1, which is seemingly the logic you're using. So you haven't actually shown the sequence converges.
songoku
Homework Statement
Find
$$\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n})$$
Relevant Equations
Limit
$$\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n})$$
$$=\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n}-n\pi)$$
$$=\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n}-n\pi)$$
$$=\lim_{n \rightarrow \infty} \sin^{2} (\pi (\sqrt{n^2+n}-n))$$
$$=\lim_{n \rightarrow \infty} \sin^{2} \left(\pi \left((\sqrt{n^2+n}-n) . \frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}\right)\right)$$
$$=\sin^{2} \left(\pi \lim_{n \rightarrow \infty} \left(\frac{n}{\sqrt{n^2+n}+n}\right)\right)$$
$$=\sin^{2} \left(\pi \left(\frac{1}{2}\right)\right)$$
$$=1$$

But if I imagine the graph of ##\sin^{2} (\pi \sqrt{n^2+n})##, it will oscillate between 0 and 1 so when ##n \rightarrow \infty##, the limit would be undefined.

Where is the mistake in my reasoning?

Thanks

songoku said:
But if I imagine the graph of ##\sin^{2} (\pi \sqrt{n^2+n})##, it will oscillate between 0 and 1
Why?

songoku
n is supposed to be an integer right? Consider what you did vs the same limit but n can be any real number. That should resolve your confusion hopefully.

songoku
PeroK said:
Why?
Office_Shredder said:
n is supposed to be an integer right? Consider what you did vs the same limit but n can be any real number. That should resolve your confusion hopefully.
I understand.

Thank you very much PeroK and Office_Shredder

Let me attempt to clarify the doubt of Songoku concretely and demonstrate that the sequence ##s_n = \sin^2 \left( \pi \sqrt{n^2 +n} \right)## converges, instead of it characteristic oscillation. So, pal, we're given
$$\sin^2 \left( \pi \sqrt{n^2 +n} \right)$$
Write ## a_n = \sqrt{n^2 +n}##. We will use, gently, (no exploitation), the fact increment in ##n## is by 1 and not less than that is possible. We will analyse the difference between consecutive ##a_n## as ##n## gets large,
$$a_{n+1} - a_n = \sqrt{ (n+1)^ + n+1} - \sqrt{ n^2 + n}$$
After rationalisation we will find
## a_{n+1} - a_n = \dfrac{ 2n+2}{ \sqrt{n^2+3n +2} + \sqrt{n^2+n} }##
We want to know what happens to this difference when ##n## gets very large, so,
$$\lim_{n \to \infty} (a_{n+1} - a_n) = \lim_{n \to \infty} \dfrac{ 2n+2}{ \sqrt{n^2+3n +2} + \sqrt{n^2+n} } = 1$$

Fix a very large ##n## and call it ##N##, such that error in ##a_{N+1} -a_N \approx 1## is beyond four decimal places, and thus, by definition of limit all subsequent ##n## will have the same property.

##\sin^2 \left( \pi a_N \right)##
## \sin^2 \left( \pi a_{N+1} \right) = \sin^2 \left( \pi a_{N}+1 \right)= \left( \sin (\pi a_N) \cos \pi + sin\pi \cos (\pi a_N) \right)^2= \sin^2 \left( \pi a_N \right)##

For any natural number ##k##,
##\sin^2 \left( \pi a_{N+k} \right) = \sin^2 \left( \pi (a_N + k) \right) = \left( sin (\pi a_N) \cos k\pi + \sin \pi \cos (\pi a_N) \right)^2 = \sin^2 \left( \pi a_N \right)##

Thus, the sequence ## \sin^2 \left( \pi \sqrt{n^2 +n} \right)## converges.

songoku said:
Homework Statement:: Find
$$\lim_{n \rightarrow \infty} \sin^{2} (\pi \sqrt{n^2+n})$$
Relevant Equations:: LimitBut if I imagine the graph of ##\sin^{2} (\pi \sqrt{n^2+n})##, it will oscillate between 0 and 1 so when ##n \rightarrow \infty##, the limit would be undefined.

Where is the mistake in my reasoning?

Thanks
Essentially, what you have shown is that ##\displaystyle \lim_{n\to \infty} \left(\sqrt{n^2+n\,} -n \right) = \frac 1 2 ## .

So, as ##n## approaches ##\infty##, ##\displaystyle \sqrt{n^2+n\,}## approaches ##\displaystyle n+ \frac 1 2 ## .

What can you say regarding ##\displaystyle \ \sin\left( \pi \left(n+ \frac 1 2 \right)\right) ##, if ##n## is even?

... if ##n## is odd?

songoku
Hall said:
Let me attempt to clarify the doubt of Songoku concretely and demonstrate that the sequence ##s_n = \sin^2 \left( \pi \sqrt{n^2 +n} \right)## converges, instead of it characteristic oscillation. So, pal, we're given
$$\sin^2 \left( \pi \sqrt{n^2 +n} \right)$$
Write ## a_n = \sqrt{n^2 +n}##. We will use, gently, (no exploitation), the fact increment in ##n## is by 1 and not less than that is possible. We will analyse the difference between consecutive ##a_n## as ##n## gets large,
$$a_{n+1} - a_n = \sqrt{ (n+1)^ + n+1} - \sqrt{ n^2 + n}$$
After rationalisation we will find
## a_{n+1} - a_n = \dfrac{ 2n+2}{ \sqrt{n^2+3n +2} + \sqrt{n^2+n} }##
We want to know what happens to this difference when ##n## gets very large, so,
$$\lim_{n \to \infty} (a_{n+1} - a_n) = \lim_{n \to \infty} \dfrac{ 2n+2}{ \sqrt{n^2+3n +2} + \sqrt{n^2+n} } = 1$$

Fix a very large ##n## and call it ##N##, such that error in ##a_{N+1} -a_N \approx 1## is beyond four decimal places, and thus, by definition of limit all subsequent ##n## will have the same property.

##\sin^2 \left( \pi a_N \right)##
## \sin^2 \left( \pi a_{N+1} \right) = \sin^2 \left( \pi a_{N}+1 \right)= \left( \sin (\pi a_N) \cos \pi + sin\pi \cos (\pi a_N) \right)^2= \sin^2 \left( \pi a_N \right)##

For any natural number ##k##,
##\sin^2 \left( \pi a_{N+k} \right) = \sin^2 \left( \pi (a_N + k) \right) = \left( sin (\pi a_N) \cos k\pi + \sin \pi \cos (\pi a_N) \right)^2 = \sin^2 \left( \pi a_N \right)##

Thus, the sequence ## \sin^2 \left( \pi \sqrt{n^2 +n} \right)## converges.
Amazingly, I think this is not a proof. If ##a_{n+1}=a_n+1+1/n## is defined recursively ,(with ##a_0=0## say) then the difference converges to 1, but I think this sequence doesn't converge if you plug it in instead of ##\sqrt{n^2+n}##

Office_Shredder said:
Amazingly, I think this is not a proof. If ##a_{n+1}=a_n+1+1/n## is defined recursively ,(with ##a_0=0## say) then the difference converges to 1, but I think this sequence doesn't converge if you plug it in instead of ##\sqrt{n^2+n}##
I didn't understand much of that.

What I did is the following:
## a_n = \sqrt{n^2 +n}##
##s_n = a_{n+1} - a_n##
##\lim s_n = 1##.

Hall said:
I didn't understand much of that.

What I did is the following:
## a_n = \sqrt{n^2 +n}##
##s_n = a_{n+1} - a_n##
##\lim s_n = 1##.
Suppose instead you had
##a_n = n + \sum_{k=1}^n 1/k##
##s_n = 1+1/n## converges to 1.

But $$\lim_{n\to \infty} \sin^2(\pi a_n)$$
Does not exist.

Last edited by a moderator:
PeroK
Office_Shredder said:
Suppose instead you had
##a_n = n + \sum_{k=1}^n 1/k##
##s_n = 1+1/n## converges to 1.

But $$\lim_{n\to \infty} \sin^2(\pi a_n)$$
Does not exist.
I realized that in post #5 in fact I proved that the sequence converges to ##\sin^2{\pi a_N}##, so if we were to change ##N## our point of convergence will change too. Upon analysing it further and contrasting it with your objection, I found that upon increasing ##N## indefinitely the difference between ##\sin^2{\pi a_N}## and ##\sin^2{\pi a_{N+k} }## goes to zero, that is the points of convergence itself converge and hence the original sequence converges; in your case we can still apply the steps of post #5 to prove that ##\sin^2{\pi a_N}## is the convergence point, but by changing ##N## the difference between convergence points doesn't reach zero. Well, all this analysis is nothing but basically Cauchy criterion.

"You're not as great as not to make any mistakes, but be as great as to accept them."

I will put myself in the second category. Thanks Office_Shredder.

## 1. What is the limit of the square of sine as x approaches infinity?

The limit of the square of sine as x approaches infinity is 1. This can be seen by using the limit definition and applying the identity sin^2(x) = 1/2 - 1/2cos(2x).

## 2. How do you find the limit of the square of sine at a specific point?

To find the limit of the square of sine at a specific point, you can use the limit definition and evaluate the function at that point. For example, if the point is x=0, then the limit is 0.

## 3. Is the limit of the square of sine always defined?

No, the limit of the square of sine is not always defined. It is only defined when the limit of sine itself is defined, which is when x approaches 0.

## 4. Can the limit of the square of sine be negative?

Yes, the limit of the square of sine can be negative. This occurs when the limit of sine itself is negative, which happens when x approaches odd multiples of pi/2.

## 5. How does the limit of the square of sine compare to the limit of sine?

The limit of the square of sine is always equal to or less than the limit of sine. This is because the square of sine is always between 0 and 1, while sine can have values greater than 1 or less than -1.

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