The expression in the image is:

$\sum_{n=2}^{\infty} \frac{1}{\sqrt{n - 1}}$

To analyze whether this series converges or diverges, we can use the comparison test or integral test due to the presence of a radical in the denominator.

Step-by-Step Solution

1. Comparison with a Known Series: Observe that $\frac{1}{\sqrt{n - 1}}$ behaves similarly to $\frac{1}{\sqrt{n}}$ for large $n$. Since we know that $\sum \frac{1}{n^{p}}$ converges if $p > 1$ and diverges if $p \leq 1$, we can see that $\frac{1}{\sqrt{n}} = \frac{1}{n^{1/2}}$ diverges, as $1/2 \leq 1$.

2. Integral Test: To confirm the divergence, consider the integral: $\int_{2}^{\infty} \frac{1}{\sqrt{x - 1}} \, dx$ By substituting $u = x - 1$, $du = dx$, the integral becomes: $\int_{1}^{\infty} \frac{1}{\sqrt{u}} \, du = \int_{1}^{\infty} u^{-\frac{1}{2}} \, du$ This integral evaluates to $\infty$, indicating divergence.

Conclusion

Since both the comparison test and integral test suggest that the series diverges, we conclude that:

$\sum_{n=2}^{\infty} \frac{1}{\sqrt{n - 1}} \text{ diverges.}$

Would you like further details or have any questions?

Here are some related questions for further exploration:

1. What are other common tests for series convergence besides the comparison and integral tests?
2. How does the p-series test apply to other values of $p$?
3. Can this series be modified in a way that would make it converge?
4. What is the difference between absolute and conditional convergence?
5. How would a summation starting at $n = 1$ instead of $n = 2$ affect convergence?

Tip: When analyzing series with radicals or polynomial terms, the comparison or integral test often gives quick insights into convergence behavior.