Let's consider the expression $\sqrt{n+1} - \sqrt{n}$ and compute the first four terms of the sequence by substituting $n = 1, 2, 3, 4$:

1. For $n = 1$: $\sqrt{1+1} - \sqrt{1} = \sqrt{2} - \sqrt{1} = \sqrt{2} - 1 \approx 1.414 - 1 = 0.414$

2. For $n = 2$: $\sqrt{2+1} - \sqrt{2} = \sqrt{3} - \sqrt{2} \approx 1.732 - 1.414 = 0.318$

3. For $n = 3$: $\sqrt{3+1} - \sqrt{3} = \sqrt{4} - \sqrt{3} = 2 - 1.732 = 0.268$

4. For $n = 4$: $\sqrt{4+1} - \sqrt{4} = \sqrt{5} - \sqrt{4} \approx 2.236 - 2 = 0.236$

Thus, the first four terms of the sequence are approximately: $0.414, \, 0.318, \, 0.268, \, 0.236$

Convergence or Divergence:

As $n \to \infty$, both $\sqrt{n+1}$ and $\sqrt{n}$ grow large, but the difference between them becomes smaller. More precisely:

$\sqrt{n+1} - \sqrt{n} = \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$

As $n \to \infty$, the denominator $\sqrt{n+1} + \sqrt{n}$ grows large, making the entire expression approach 0. Therefore, the sequence converges to 0.

Conclusion:

The sequence is convergent.

Would you like further details or have any questions?

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Here are 5 related questions to expand your understanding:

1. What is the general form of the limit for sequences involving square roots?
2. How can you prove the convergence of a sequence using mathematical limits?
3. What happens if you try this with a different form, like $\sqrt{n+2} - \sqrt{n}$?
4. How would the sequence behave if $n$ starts at 0 instead of 1?
5. How can we approximate the difference $\sqrt{n+1} - \sqrt{n}$ for large $n$?

Tip: In sequences involving square roots, rationalizing the expression often helps in determining convergence properties!