## University Calculus: Early Transcendentals (3rd Edition)

We are given that $a_{n+1}=(a_n)^{n+1}$ and $a_1=\dfrac{1}{2}$ Here, for $n=1,2,3,....n$ we have $a_2=(a_1)^{1+1}=(\dfrac{1}{2})^{2} \\ a_3=(a_2)^{2+1}=(\dfrac{1}{2})^{6}.... \\ ...a_n=(\dfrac{1}{2})^{n!}$ Now apply the direct comparison test: $\Sigma_{n=1}^\infty (\dfrac{1}{2})^{n!} \leq \Sigma_{n=1}^\infty(\dfrac{1}{2})^n$ Here, the series $\Sigma_{n=1}^\infty(\dfrac{1}{2})^n$ converges Hence, the given series converges.