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If $$\large{a_n = \frac{x^n}{2^n n!}}$$ , Then find $$\large{ \frac{a_{n +1}}{a_n}}$$

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I tried the following: $$\large{a_{n + 1} = \frac{x^n}{2^n n!} + \frac{2^n n!}{ 2^n n!} = \frac{x^n + 2^n n!}{2^n n!}}$$

Then I divided this thing by an and got: $$\large{x^n(x^n + 2^n n!) = 3x^{2n} n!}$$

But there was no such choice where I faced this question so I got it wrong. How could I really solve this?

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  • $\begingroup$ Please check if my edit of $a_{n+1}$ is correct, or if it was $a_n+1$ $\endgroup$ – Vladimir Vargas Jun 19 '15 at 14:43
  • $\begingroup$ It is correct. I think I just didn't understand the question correctly at first. $\endgroup$ – m1k2zoo Jun 19 '15 at 14:56
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Note $a_n = \frac{x^n}{2^n \cdot n!}$. So we have:

$$a_{n+1} = \frac{x^{n+1}}{2^{n+1} \cdot (n+1)!}$$

Thus we see:

$$\frac{a_{n+1}}{a_n} = \frac{x^{n+1}}{2^{n+1} \cdot (n+1)!} \cdot \frac{2^n \cdot n!}{x^n} $$

Can you simplify further?

EDIT:

Note that $n! = 1 \cdot 2 \cdot 3 \: \cdot \: ... \: \cdot \: (n-1) \cdot n$.

So that means that $(n+1)! = 1 \cdot 2 \: \cdot \: ... \: \cdot \: (n-1) \cdot n \cdot (n+1)$.

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  • $\begingroup$ By exponential rules: $\endgroup$ – m1k2zoo Jun 19 '15 at 14:56
  • $\begingroup$ I got x/2. But I'm not sure about simplifying the factorials. $\endgroup$ – m1k2zoo Jun 19 '15 at 14:58
  • $\begingroup$ @m1k2zoo See my edit, explaining factorials. Does that help? $\endgroup$ – anakhro Jun 19 '15 at 15:06
  • $\begingroup$ So we can say that (n+1)!=n!(n+1). Hence the n! at the top will go with the n! in the denominator and we'll be lefting with (n+1). $\endgroup$ – m1k2zoo Jun 19 '15 at 15:16
  • $\begingroup$ It does really make sense now. Thank you. $\endgroup$ – m1k2zoo Jun 19 '15 at 15:17
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$$\frac{a_{n+1}}{a_n}=\frac{\frac{x^{n+1}}{2^{n+1}(n+1)!}}{\frac{x^n}{2^nn!}}=\frac{x^{n+1}}{2^{n+1}(n+1)!}\frac{2^nn!}{x^n}=\frac{x}{2(n+1)}$$

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