سوال ۳۳ کنکور ارشد علوم کامپیوتر ۱۴۰۱

Simplify the numerator. Using the angle-addition identity,

$\sin\!\left(\left(n+\tfrac{1}{2}\right)\pi\right) = \sin\!\left(n\pi + \tfrac{\pi}{2}\right) = \cos(n\pi) = (-1)^{n}.$

So the series is exactly an alternating series:

$\sum_{n=10}^{\infty} \frac{(-1)^{n}}{\ln(\ln n)}.$

Convergence (Leibniz test). Let $b_n = \dfrac{1}{\ln(\ln n)}$. For $n \ge 10$ we have $\ln n > 1$, so $\ln(\ln n) > 0$ and is strictly increasing to $\infty$. Hence $b_n > 0$, $b_n$ is decreasing, and $b_n \to 0$. By the alternating series (Leibniz) test, the series converges.

Not absolute convergence. The series of absolute values is $\sum \dfrac{1}{\ln(\ln n)}$. Since $\ln(\ln n)$ grows extremely slowly, $\ln(\ln n) < n$ for all large $n$, giving

$\frac{1}{\ln(\ln n)} > \frac{1}{n}.$

Because the harmonic series $\sum \tfrac{1}{n}$ diverges, the comparison test shows $\sum \dfrac{1}{\ln(\ln n)}$ diverges. (Equivalently, the terms do not even shrink fast enough — they dominate $1/n$.)

Conclusion. The series converges but not absolutely, so it is conditionally convergent.

Exam trick: Whenever you see $\sin\big((n+\tfrac12)\pi\big)$ or $\cos(n\pi)$, replace it immediately with $(-1)^n$. The whole problem collapses to the Leibniz test plus an absolute-value check.