سوال ۳۲ کنکور ارشد علوم کامپیوتر ۱۳۹۸

We need the set of exponents $p>0$ for which $\sum_{n=1}^{\infty}\left(\sqrt[n]{n}-1\right)^{p}$ converges. Everything is controlled by the asymptotic size of the base term $a_n = \sqrt[n]{n}-1$.

Asymptotics of $\sqrt[n]{n}-1$

Write $\sqrt[n]{n} = n^{1/n} = e^{(\ln n)/n}$. As $n\to\infty$ the exponent $\tfrac{\ln n}{n}\to 0$, so by $e^{x}-1 \sim x$ we get

$\sqrt[n]{n}-1 = e^{(\ln n)/n}-1 \sim \frac{\ln n}{n}\qquad (n\to\infty).$

Hence the general term behaves like

$\left(\sqrt[n]{n}-1\right)^{p} \sim \left(\frac{\ln n}{n}\right)^{p} = \frac{(\ln n)^{p}}{n^{p}}.$

Convergence test

The convergence of $\sum a_n^{p}$ is the same as that of $\sum \dfrac{(\ln n)^{p}}{n^{p}}$ (limit comparison, the ratio tends to a positive constant). This is a $p$-series with a logarithmic factor:

• If $p>1$: pick any $q$ with $1<q<p$. Since $(\ln n)^{p} = o\!\left(n^{\,p-q}\right)$, we have $\dfrac{(\ln n)^{p}}{n^{p}} \le \dfrac{1}{n^{q}}$ for large $n$, and $\sum n^{-q}$ converges. So the series converges.
• If $p=1$: the term is $\sim \dfrac{\ln n}{n} \ge \dfrac{1}{n}$ for $n\ge 3$, and $\sum \tfrac{1}{n}$ diverges. So it diverges.
• If $0<p<1$: $\dfrac{(\ln n)^{p}}{n^{p}} \ge \dfrac{1}{n^{p}}$ eventually (since $\ln n\ge 1$), and $\sum n^{-p}$ diverges because $p<1$. So it diverges.

Conclusion

Convergence holds exactly when $p>1$, so

$A = (1,\infty).$

The boundary $p=1$ is excluded (the logarithm pushes it just past the harmonic series into divergence), which is why the interval is open. This is option 1.

Pitfall: Don't be tempted to include $p=1$ (option 2). The bare comparison $a_n \sim 1/n$ might suggest a borderline, but the extra $\ln n$ factor makes $\sum \frac{\ln n}{n}$ strictly larger than the harmonic series, so $p=1$ diverges and the interval is open.