## Thomas' Calculus 13th Edition

Given: $a_{n+1}=(a_n)^{n+1}$ and $a_1=\dfrac{1}{2}$; and $n=1,2,3,....n$ Here, we can see that $a_2=(a_1)^{1+1}=(\dfrac{1}{2})^{2} \\ a_3=(a_2)^{2+1}=(\dfrac{1}{2})^{3}.....$ and $a_n=(\dfrac{1}{2})^{n!}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}(\dfrac{1}{2})^{n!}$ Next, we need to apply the direct comparison test. $\Sigma_{n=1}^\infty (\dfrac{1}{2})^{n!} \leq \Sigma_{n=1}^\infty(\dfrac{1}{2})^n$ we can see that the series $\Sigma_{n=1}^\infty(\dfrac{1}{2})^n$ converges Therefore, the series converges.