Strategy for Series — Question 4

Question 4 – Solution

Step 1: Try the natural factorial test.

If $a_n=(2n)!/[(n!)^2 4^n]$, then $$\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{4(n+1)^2}=\frac{2n+1}{2n+2}\longrightarrow1.$$ The Ratio Test is therefore inconclusive.

Step 2: Replace it with asymptotic comparison.

Stirling’s formula yields the standard estimate $$\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}.$$ Consequently, $$a_n=\frac1{4^n}\binom{2n}{n}\sim\frac1{\sqrt{\pi n}}.$$ Equivalently, $a_n/(1/\sqrt n)\to1/\sqrt\pi>0$.

Conclusion.

Since $\sum n^{-1/2}$ diverges, the given series diverges by Limit Comparison. A ratio limit of $1$ signaled that a finer growth estimate was needed.