سوال ۲۶ کنکور ارشد مهندسی کامپیوتر ۱۴۰۴

Write the general term using $\ln\dfrac{n}{n+2} = \ln n - \ln(n+2)$, so

$a_n = (-1)^n\big(\ln n - \ln(n+2)\big).$

The series is alternating, $\left|\ln\frac{n}{n+2}\right|\to 0$ monotonically, so by the Leibniz test it converges. To find the sum, compute the partial sum $S_N$ for even $N$ by tracking the coefficient of each $\ln k$.

The $\ln n$ part gives $\sum_{n=1}^{N}(-1)^n\ln n$, and the $-\ln(n+2)$ part gives

$-\sum_{n=1}^{N}(-1)^n\ln(n+2) = -\sum_{m=3}^{N+2}(-1)^m\ln m.$

The shared range $m=3,\dots,N$ cancels (interior terms telescope), leaving only the boundary:

$S_N = \underbrace{\big((-1)^1\ln 1 + (-1)^2\ln 2\big)}_{=\,\ln 2} \;-\;\big((-1)^{N+1}\ln(N+1)+(-1)^{N+2}\ln(N+2)\big).$

For even $N$ the boundary term is $\ln(N+1)-\ln(N+2)=\ln\dfrac{N+1}{N+2}$, hence

$S_N = \ln 2 + \ln\frac{N+1}{N+2}.$

Letting $N\to\infty$, $\ln\dfrac{N+1}{N+2}\to\ln 1 = 0$, so

$\sum_{n=1}^{\infty} (-1)^n \ln\frac{n}{n+2} = \ln 2.$

The answer is option 4.

Exam trick: Split $\ln\frac{n}{n+2}$ into $\ln n-\ln(n+2)$ and note every interior $\ln k$ cancels — only the first two and last two terms of the partial sum survive, so the limit is read off the boundary alone.