To prove that the sequence $\{a_n\}$ converges to a limit $\alpha$, we need to:

1. Identify the limit $\alpha$.
2. Use the definition of the limit of a sequence: For every $\epsilon > 0$, we need to find $N(\epsilon) \in \mathbb{N}$ such that for all $n \geq N(\epsilon)$, $|a_n - \alpha| < \epsilon$.

Sequence: $a_n = \frac{\sqrt{n}}{n^2 + 1}$

Step 1: Find the Limit $\alpha$

We will first analyze the asymptotic behavior of the sequence $a_n$ as $n \to \infty$.

$a_n = \frac{\sqrt{n}}{n^2 + 1}$

For large $n$, we can approximate the denominator $n^2 + 1$ by $n^2$. So, for large $n$, we have:

$a_n \approx \frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}}$

As $n \to \infty$, $\frac{1}{n^{3/2}} \to 0$. Hence, the limit of $a_n$ as $n \to \infty$ is $\alpha = 0$.

Step 2: Definition of Limit

We want to show that for every $\epsilon > 0$, there exists $N(\epsilon) \in \mathbb{N}$ such that for all $n \geq N(\epsilon)$:

$|a_n - 0| = \left| \frac{\sqrt{n}}{n^2 + 1} \right| < \epsilon$

This simplifies to:

$\frac{\sqrt{n}}{n^2 + 1} < \epsilon$

Step 3: Solve for $N(\epsilon)$

To find $N(\epsilon)$, we start by simplifying the inequality:

$\frac{\sqrt{n}}{n^2 + 1} < \epsilon$

Since $n^2 + 1 > n^2$, we have the bound:

$\frac{\sqrt{n}}{n^2 + 1} < \frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}}$

Thus, we want:

$\frac{1}{n^{3/2}} < \epsilon$

Taking the reciprocal of both sides (and flipping the inequality):

$n^{3/2} > \frac{1}{\epsilon}$

Now, solve for $n$:

$n > \left( \frac{1}{\epsilon} \right)^{2/3}$

Thus, we can take:

$N(\epsilon) = \left\lceil \left( \frac{1}{\epsilon} \right)^{2/3} \right\rceil$

This means that for any $\epsilon > 0$, if $n \geq N(\epsilon)$, we have $\left| \frac{\sqrt{n}}{n^2 + 1} \right| < \epsilon$, which proves that the sequence converges to $\alpha = 0$.

Conclusion

The sequence $\{a_n\}$ converges to $\alpha = 0$, and for any $\epsilon > 0$, $N(\epsilon) = \left\lceil \left( \frac{1}{\epsilon} \right)^{2/3} \right\rceil$ satisfies the definition of the limit.

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Would you like further details, or do you have any questions?

Related Questions:

1. How would the sequence change if the exponent of $n$ in the denominator were different?
2. Can this approach be generalized to other sequences involving square roots and polynomials?
3. What is the difference between using $n^2 + 1$ versus $n^2$ in the denominator for large $n$?
4. What other sequences have limits of zero, and how can we recognize them quickly?
5. How does this proof change if the sequence had a different power of $n$ in the numerator?

Tip:

When dealing with limits of sequences, approximating the behavior of the terms for large $n$ often simplifies the process of finding the limit.