Show that each of the following sequences {an} converges to a limit α (say). For given ϵ > 0, find an
N(ϵ) ∈ N as required in the definition of limit.
(i) an =root(n)/n^2+1

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.

---

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.

Show that each of the following sequences {an} converges to a limit α (say). For given ϵ > 0, find an N(ϵ) ∈ N as required in the definition of limit. (i) an = sqrt(n)/(n^2 + 1)

This article shows how to find the limit of the sequence a_n = sqrt(n)/(n^2 + 1) as n approaches infinity. It includes step-by-step calculations and the definition of limit, focusing on finding N(ϵ) for any given ϵ > 0. Suitable for undergraduate students learning about sequences and limits.

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