سوال ۳۲ کنکور ارشد علوم کامپیوتر ۱۳۹۷

Step 1 — Find a closed form for $a_n$.

Near $x = 0$, the Taylor expansion gives $1 - \cos x = \tfrac{x^{2}}{2} + O(x^{4})$, so $1 - \cos x \sim \frac{x^{2}}{2} \quad (x \to 0).$ Raising to the power $\tfrac{n}{2}$ (the quantity is positive, so the fractional power is well defined): $(1-\cos x)^{n/2} \sim \left(\frac{x^{2}}{2}\right)^{n/2} = \frac{(x^{2})^{n/2}}{2^{n/2}} = \frac{|x|^{n}}{2^{n/2}}.$

Step 2 — Handle the sign from $x \to 0^-$.

The denominator is $x^{n}$. For $x < 0$ we have $x^{n} = (-1)^{n}|x|^{n}$, hence $\frac{(1-\cos x)^{n/2}}{x^{n}} \sim \frac{|x|^{n}}{2^{n/2}\,(-1)^{n}|x|^{n}} = \frac{(-1)^{n}}{2^{n/2}} = \left(-\frac{1}{\sqrt{2}}\right)^{n}.$ The $|x|^{n}$ cancels exactly, so the limit exists and $a_n = \left(-\frac{1}{\sqrt{2}}\right)^{n}.$

Step 3 — Analyse the series.

Thus $\sum_{n=1}^{\infty} a_n$ is a geometric series with ratio $r = -\frac{1}{\sqrt{2}}$ and $|r| = \frac{1}{\sqrt{2}} < 1$. The series of absolute values $\sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{\infty} \left(\frac{1}{\sqrt{2}}\right)^{n} = \frac{1/\sqrt{2}}{1 - 1/\sqrt{2}}$ is a convergent geometric series, so $\sum a_n$ is absolutely convergent.

Checking the options:

• Option 1: $a_n = (-1/\sqrt{2})^{n} \to 0$, so $\lim a_n \neq 0$ is false.
• Option 2: absolutely convergent — true.
• Option 3: conditionally convergent is false (absolute convergence is stronger, and it holds).
• Option 4: the series converges, so "divergent" is false.

The correct choice is option 2.

Exam trick: $1-\cos x \sim x^{2}/2$ turns the fractional power into a clean power of $|x|$; the only subtlety is the side $x\to 0^-$, which injects the alternating sign $(-1)^n$ but does not affect absolute convergence.