Limit of sin(π sqrt(n² + 1))
Evaluate the limit: $$ L = \lim_{n \to \infty} \sin\left(\pi\sqrt{n^2+1}\right), \qquad n \text{ a positive integer.} $$
Explanation of Method
Since $\sqrt{n^2+1}$ is irrational for every integer $n\ge1$ and grows without bound, $\sin\left(\pi\sqrt{n^2+1}\right)$ does not obviously settle down on its own. The key is to isolate exactly how far $\sqrt{n^2+1}$ sits from the nearby integer $n$: a binomial approximation shows $\sqrt{n^2+1}=n+\frac1{2n}+O(n^{-3})$, so the sine's argument is $\pi n$ plus a small correction. Since $\sin(\pi n+\theta)=(-1)^n\sin\theta$ exactly (not simply $\sin\theta$ — the sign alternates with the parity of $n$), care is needed with that alternating sign; but since $|\sin\theta|\le|\theta|$, the whole expression is squeezed to $0$ regardless of the sign.
- Expand $\sqrt{n^2+1}$ using the binomial approximation for large $n$.
- Rewrite the sine using the exact identity $\sin(\pi n+\theta)=(-1)^n\sin\theta$.
- Bound the result using $|\sin\theta|\le|\theta|$ and apply the Squeeze Theorem.
Step 1 · Expand $\sqrt{n^2+1}$ for large $n$
Write $\sqrt{n^2+1}=n\sqrt{1+\frac1{n^2}}$, and use the binomial approximation $\sqrt{1+u}\approx1+\frac u2$ for small $u=\frac1{n^2}$:
$$ \sqrt{n^2+1} = n\left(1+\frac1{2n^2}+O(n^{-4})\right) = n+\frac1{2n}+O(n^{-3}). $$
Step 2 · Rewrite the sine using the exact periodicity identity
Let $\theta = \pi\left(\frac1{2n}+O(n^{-3})\right)$, so that $\pi\sqrt{n^2+1}=\pi n+\theta$. Using the exact identity (from the angle-addition formula, since $\sin(\pi n)=0$ and $\cos(\pi n)=(-1)^n$ for every integer $n$):
$$ \sin(\pi n+\theta) = \sin(\pi n)\cos\theta+\cos(\pi n)\sin\theta = 0\cdot\cos\theta+(-1)^n\sin\theta = (-1)^n\sin\theta. $$
So:
$$ \sin\left(\pi\sqrt{n^2+1}\right) = (-1)^n\sin\left(\frac{\pi}{2n}+O(n^{-3})\right). $$
Step 3 · Bound and apply the Squeeze Theorem
Using $|\sin\theta|\le|\theta|$ for every real $\theta$:
$$ \left|\sin\left(\pi\sqrt{n^2+1}\right)\right| = \left|\sin\left(\frac{\pi}{2n}+O(n^{-3})\right)\right| \le \frac{\pi}{2n}+O(n^{-3}). $$
As $n\to\infty$, the right-hand side tends to $0$. By the Squeeze Theorem (applied to the absolute value), the original expression is squeezed to $0$ regardless of the alternating sign $(-1)^n$:
$$ L = \lim_{n\to\infty}\sin\left(\pi\sqrt{n^2+1}\right) = 0. $$
Summary Table
| Quantity | Behavior as $n\to\infty$ |
|---|---|
| $\sqrt{n^2+1}-n$ | $\dfrac1{2n}+O(n^{-3}) \to 0$ |
| $\sin\left(\pi\sqrt{n^2+1}\right)$ | $(-1)^n\sin\left(\dfrac\pi{2n}+O(n^{-3})\right)$ |
| $\left|\sin\left(\pi\sqrt{n^2+1}\right)\right|$ | $\le\dfrac\pi{2n}+O(n^{-3}) \to 0$ |
Additional Notes
- The identity $\sin(\pi n+\theta)=(-1)^n\sin\theta$ (rather than simply $\sin\theta$) is essential to state correctly: for even $n$ the two agree, but for odd $n$ they differ by a sign — since the goal is a limit of $0$, and both $+\sin\theta\to0$ and $-\sin\theta\to0$ as $\theta\to0$, the alternating sign turns out not to affect the final conclusion, but it would matter if the problem instead asked for the limit of, say, $(-1)^n\sin\left(\pi\sqrt{n^2+1}\right)$.
- Working with the absolute value and the Squeeze Theorem sidesteps the need to track the sign $(-1)^n$ explicitly — bounding $|\sin(\pi\sqrt{n^2+1})|$ above by a quantity tending to $0$ is enough to pin down the limit of the original (signed) sequence as well, since $-|\cdot| \le \sin(\cdot) \le |\cdot|$ always.
- This problem is a good illustration of why sequences involving $\sin(\pi\cdot f(n))$ for $f(n)$ close to an integer should always be handled via the exact distance $f(n)-\lfloor f(n)+\tfrac12\rfloor$ (or, as here, $f(n)-n$) rather than assumed periodicity shortcuts that silently drop a sign.
Result
The value of the limit is: