91Ó°ÊÓ

First, we have to identify the terms of the sequence. The given series is: $$\frac{1}{2^{2}}+\frac{2}{3^{2}}+\frac{3}{4^{2}}+\cdots$$ This can be written as: $$\sum_{n=1}^{\infty} \frac{n}{(n+1)^2}$$

We need to find a convergent or divergent sequence to compare our sequence with. Let's compare it with the convergent sequence \(\frac{1}{n(n+1)}\). Therefore, the comparison is: $$\frac{n}{(n+1)^2} \text{ and } \frac{1}{n(n+1)}$$

Now, let's check if the following inequality holds for all \(n \ge 1\): $$0 \le \frac{1}{n(n+1)} \le \frac{n}{(n+1)^2}$$ Since the numerator is less than or equal to the denominator for both fractions, the inequality holds.

Since we are using the Comparison Test, we need to confirm convergence of the compared sequence. The sum of the sequence \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\) can be found by partial fraction decomposition: $$\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$$ Multiplying both sides by \(n(n+1)\), we get: $$1 = A(n+1) + Bn$$ Comparing coefficients, we have: - For \(A\): \(1 = A(1+1)\) ⇒ \(A = \frac{1}{2}\) - For \(B\): \(1 = B(1)\) ⇒ \(B = 1\) Substituting values of \(A\) and \(B\) in the partial fraction decomposition, we get: $$\frac{1}{n(n+1)} = \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+1}\right)$$ Now we can compute the sum of the series: $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{2} \cdot \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)$$ Now using the telescoping method, the series will collapse and: $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{2} (1)$$ Since the compared sequence converges, we can conclude that our original sequence converges as well.

By the Comparison Test, the original series $$\sum_{n=1}^{\infty} \frac{n}{(n+1)^2}$$ converges.