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I have this series:

$$a_n = \frac{(n!)^2}{2^{n^2}}$$

I tried to solve it with:

$$\lim_{n\to\infty} \frac{a_n+1}{a_n}$$

So I get:

$$\lim_{n\to\infty} \frac{\frac{\big[(n+1)!\big]^2}{2^{(n+1)^2}}}{\frac{(n!)^2}{2^{n^2}}}$$ $$\lim_{n\to\infty} \frac{\big[(n+1)!\big]^2}{2^{(n+1)^2}} * \frac{2^{n^2}}{(n!)^2}$$ $$\lim_{n\to\infty} \frac{\big[(n+1)(n)!\big]^2}{2^{(n+1)^2}} * \frac{2^{n^2}}{(n!)^2}$$ $$\lim_{n\to\infty} \frac{\big[(n+1)\big]^2}{2^{(n+1)^2}} * 2^{n^2}$$ $$\lim_{n\to\infty} \frac{2^{n^2}(n+1)^2}{2^{(n+1)^2}}$$

How should I proceed?

GoodDeeds
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Christian Giupponi
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2 Answers2

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After this, using the laws of indices, $$\lim_{n\rightarrow\infty}\frac{2^{n^2}(n+1)^2}{2^{(n+1)^2}}=\lim_{n\rightarrow\infty}\frac{(n+1)^2}{2^{(n+1)^2-n^2}}=\lim_{n\rightarrow\infty}\frac{(n+1)^2}{2^{2n+1}}=0$$

Thus, the series is convergent.

GoodDeeds
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Going from $n$ to $n+1$, you multiply by $\dfrac{(n+1)!^2}{n!^2}=(n+1)^2$ and divide by $\dfrac{2^{(n+1)^2}}{2^{n^2}}=2^{2n+1}$. The denominator "wins". (For all $n$, the ratio does not exceed $\dfrac12$.)