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Solution:

Since, f(n) = \(\frac{1}{1 + n^2}\) which is continuous, positive and decreasing for n >= 0

Here,

\(\int_{0}^{∞} (\frac{1}{1 + x^2})dx \)

\(= \lim_{k \to ∞} \int_{0}^{k} \frac{dx}{1 + x^2}\)

\(= \lim_{k \to ∞} [tan^{-1}x]_{0}^{k}\)

\(= \lim_{k \to ∞} [tan^{-1}(k) – tan^{-1}(0)]\)

\(= \frac{\pi}{2}\)

This means the integral converges. Therefore, by integral test, the given test is convergent.

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