How do I use the ratio test to determine convergence or divergence in this problem?

Question

I have the problem:

$$a_{n} = \frac{e^{n+5}}{\sqrt{n+7}(n+3)!}$$

I am told to use the ratio test to determine convergence or divergence (or the test could be inconclusive).

So I take the limit:

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

$$\lim_{n\to \infty} \frac{e^{n+6}}{\sqrt{n+8}(n+4)n!} \frac{\sqrt{n+7}(n+3)n!}{e^{n+5}}$$ $$= (\lim_{n\to \infty} \frac{e^{n+6}}{e^{n+5}}) (\lim_{n\to \infty} \frac{\sqrt{n+7}}{\sqrt{n+8}}) (\lim_{n\to \infty} \frac{n+3}{n+4})$$

I reorganized and canceled the n! terms.

I keep getting: $$(\lim_{n\to \infty} \frac{e^{n+6}}{e^{n+5}}) = e$$ $$(\lim_{n\to \infty} \frac{\sqrt{n+7}}{\sqrt{n+8}}) = 1$$ $$(\lim_{n\to \infty} \frac{n+3}{n+4}) = 1$$

So my limit is e, which indicates the series diverges. However, according both to MathWorks and WolframAlpha this is incorrect. The limit is actually 0, indicating the series converges.

Where am I going wrong?

Answer 1

You made a mistake writing the ratio. It should be $$ \frac{e^{n+6}}{\sqrt{n+8}(n+4)!} \frac{\sqrt{n+7}(n+3)!}{e^{n+5}} = \frac{e\cdot e^{n+5}}{\sqrt{n+8}(n+4)(n+3)!} \frac{\sqrt{n+7}(n+3)!}{e^{n+5}} $$ not $$\frac{e^{n+6}}{\sqrt{n+8}(n+4)n!} \frac{\sqrt{n+7}(n+3)n!}{e^{n+5}}.$$

The mistake is in that you wrote $(n+k)! = (n+k)n!$, which is not true.

Answer 2

Note that

$$\dfrac{a_{n+1}}{a_n}=\dfrac{\dfrac{e^{n+6}}{(n+4)!\sqrt{n+8}}}{\dfrac{e^{n+5}}{(n+3)!\sqrt{n+7}}}=\dfrac{e\sqrt{n+7}}{(n+4)\sqrt{n+8}}\underbrace{\to}_{n\to \infty} 0.$$