Convergence (2)

$\begin{aligned} \sum_{n=0}^\infty a_n & = \sum_{n=0}^\infty \frac {11^n((kn)!)^2}{(24n)!} & \small \color{#3D99F6} \text{Stirling's formula: }n! \approx \sqrt {2\pi} n^{n+1/2} e^{-n} \\ & = \sum_{n=0}^\infty \frac {11^n \sqrt {2\pi} (kn)^{2kn+1}e^{-2kn}}{(24n)^{24n+1/2} e^{-24n}} \end{aligned}$

Consider the ratio text, for the sum to converge, we have $\displaystyle L = \lim_{n \to \infty} \left| \frac {a_{n+1}}{a_n} \right| < 1$ . Therefore,

$\begin{aligned} L & = \lim_{n \to \infty} \frac {11^{n+1} \sqrt {2\pi}(k(n+1))^{2k(n+1)+1}e^ {-2k(n+1)}}{(24(n+1))^{24(n+1)+1/2} e^{-24(n+1)}} \cdot \frac {(24n)^{24n+1/2} e^{-24n}}{11^n \sqrt {2\pi}(kn)^{2kn+1}e^{-2kn}} \\ & = \frac {11(kn)^{2k}e^{24}}{(24n)^{24} e^{2k}} \end{aligned}$

We note that when $k=13$ , $L \to \infty$ . For $L < 1$ , the largest $k_{max} = \boxed{12}$ .

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References:

• Stirling's formula
• Ratio test