The sum of the series as $n \to \infty$ $\frac{\sqrt{n}}{(3+4\sqrt{n})^2} + \frac{\sqrt{n}}{\sqrt{2}(3\sqrt{2}+4\sqrt{n})^2} + \frac{\sqrt{n}}{\sqrt{3}(3\sqrt{3}+4\sqrt{n})^2} + \ldots + \frac{1}{49n}$ is:
Step-by-Step Solution
Key Concept: Convert the series into a Riemann sum by factoring out √n and recognizing the general term as √n/(k√k · (3√k + 4√n)²). As n→∞, normalize by setting x = k/n to transform the sum into an integral ∫₀¹ 1/(√x(3√x + 4)²) dx using substitution 3√x + 4 = t.
The limit $P = \lim_{n→∞} \sum_{k=1}^{n} \frac{\sqrt{n}}{k\sqrt{3\sqrt{n}+4\sqrt{n}}}$ is rewritten as a Riemann sum. Substituting $3\sqrt{x} + 4 = t$ gives $\frac{3}{2\sqrt{x}}dx = dt$. The integral becomes $P = \int_4^7 \frac{dt}{3t^2} = \frac{1}{3}\left[-\frac{1}{t}\right]_4^7 = \frac{2}{3}\left(\frac{1}{4} - \frac{1}{7}\right)$.
Correct Answer: 1