Question

The value of
\[ \lim_{n \to \infty} \sum_{k=2}^n \frac{\sqrt{n} + 1 - \sqrt{n}}{k(\ln k)^2} \]
is equal to

• A. \(\infty\)
• B. \(1\)
• C. \(e\)
• D. \(0\)

Answer 1

The Correct Option is D

To evaluate the limit, we start by considering the expression inside the summation:

\[\frac{\sqrt{n} + 1 - \sqrt{n}}{k(\ln k)^2}\]

Simplifying the numerator:

\[\sqrt{n} + 1 - \sqrt{n} = 1\]

The expression becomes:

\[\frac{1}{k (\ln k)^2}\]

Thus, the original limit can be rewritten as:

\[\lim_{n \to \infty} \sum_{k=2}^n \frac{1}{k (\ln k)^2}\]

The sum:

\[\sum_{k=2}^n \frac{1}{k (\ln k)^2}\]

is a series that can be compared integrally. To understand its behavior as \(n \to \infty\), consider the integral:

\[\int_2^n \frac{1}{x (\ln x)^2} \, dx\]

This integral can be evaluated by the substitution method. Let \(u = \ln x\), then \(du = \frac{1}{x} \, dx\). Thus, the integral becomes:

\[\int_{u=\ln 2}^{u=\ln n} \frac{1}{u^2} \, du = \left[-\frac{1}{u}\right]_{\ln 2}^{\ln n}\]

Evaluating the limits:

\[-\frac{1}{\ln n} + \frac{1}{\ln 2}\]

As \(n \to \infty\), \(\ln n \to \infty\), therefore:

\[-\frac{1}{\ln n} \to 0\]

This implies the integral:

\[\frac{1}{\ln 2} = \text{constant as } n \to \infty\]

Thus, the series converges, and the limit of the series:

\[\sum_{k=2}^n \frac{1}{k (\ln k)^2}\]

converges to 0 as \(n \to \infty\) since its partial sums approach a finite value.

Hence, the value of:

\[\lim_{n \to \infty} \sum_{k=2}^n \frac{\sqrt{n} + 1 - \sqrt{n}}{k(\ln k)^2}\]

is:

\(0\)