Question

If $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left( \frac{n^2 + (k-1)^2}{n^2 + k^2} \right)$ exists and is equal to L. The absolute value of [L] is (where [.] denotes greatest integer function)

Answer 1

0

Answer 2

The given limit is $L = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left( \frac{n^2 + (k-1)^2}{n^2 + k^2} \right)$.

We can use the property of logarithms $\ln(a/b) = \ln a - \ln b$ to rewrite the term inside the sum: $\ln \left( \frac{n^2 + (k-1)^2}{n^2 + k^2} \right) = \ln(n^2 + (k-1)^2) - \ln(n^2 + k^2)$.

Let $f(k) = \ln(n^2 + k^2)$. Then the term in the sum is $f(k-1) - f(k)$. The sum becomes $\sum_{k=1}^{n} [f(k-1) - f(k)]$. This is a telescoping sum: $\sum_{k=1}^{n} [f(k-1) - f(k)] = [f(0) - f(1)] + [f(1) - f(2)] + \dots + [f(n-1) - f(n)]$ $= f(0) - f(n)$.

Now we evaluate $f(0)$ and $f(n)$: $f(0) = \ln(n^2 + 0^2) = \ln(n^2)$. $f(n) = \ln(n^2 + n^2) = \ln(2n^2)$.

The sum is $f(0) - f(n) = \ln(n^2) - \ln(2n^2) = \ln\left(\frac{n^2}{2n^2}\right) = \ln\left(\frac{1}{2}\right) = -\ln 2$.

The expression inside the limit is $\frac{1}{n} \sum_{k=1}^{n} [\ln(n^2 + (k-1)^2) - \ln(n^2 + k^2)] = \frac{1}{n} (-\ln 2) = -\frac{\ln 2}{n}$.

Now we compute the limit $L$: $L = \lim_{n \to \infty} -\frac{\ln 2}{n}$. Since $\ln 2$ is a constant, and $\lim_{n \to \infty} \frac{1}{n} = 0$, we have: $L = -(\ln 2) \times \lim_{n \to \infty} \frac{1}{n} = -(\ln 2) \times 0 = 0$.

So, $L = 0$. The question asks for the absolute value of $[L]$, where $[.]$ denotes the greatest integer function. $[L] = [0] = 0$. The absolute value of $[L]$ is $|[L]| = |0| = 0$.