Eliza Gregory

2022-10-13

Determine whether the Sequence is decreasing or increasing.

I have the sequence $\frac{({10}^{n})}{(2n)!}$ and am trying to determine whether the sequence decreases or increases. I feel like the best way to proceed would be to use the squeeze theorem, but am unsure how to apply it to the problem.

I have the sequence $\frac{({10}^{n})}{(2n)!}$ and am trying to determine whether the sequence decreases or increases. I feel like the best way to proceed would be to use the squeeze theorem, but am unsure how to apply it to the problem.

Martha Dickson

Beginner2022-10-14Added 20 answers

The hint:

$$\frac{{a}_{n+1}}{{a}_{n}}=\frac{\frac{{10}^{n+1}}{(2n+2)!}}{\frac{{10}^{n}}{(2n)!}}=\frac{5}{(n+1)(2n+1)}<1$$

for all $n\ge 1$

$$\frac{{a}_{n+1}}{{a}_{n}}=\frac{\frac{{10}^{n+1}}{(2n+2)!}}{\frac{{10}^{n}}{(2n)!}}=\frac{5}{(n+1)(2n+1)}<1$$

for all $n\ge 1$

Ignacio Riggs

Beginner2022-10-15Added 4 answers

Let ${a}_{n}=\frac{{10}^{n}}{(2n)!}$ defined for natural numbers. Now, let examine the following ratio

$$\frac{{a}_{n+1}}{{a}_{n}}=\frac{\frac{{10}^{n+1}}{(2(n+1))!}}{\frac{{10}^{n}}{(2n)!}}=\frac{10}{(2n+1)(2n+2)}$$

Now, observe that $\frac{{a}_{n+1}}{{a}_{n}}<1$ for every $n\ge 1$. Thus, ${a}_{n+1}<{a}_{n}$ which means that $\{{a}_{n}\}$ is decreasing sequence.

$$\frac{{a}_{n+1}}{{a}_{n}}=\frac{\frac{{10}^{n+1}}{(2(n+1))!}}{\frac{{10}^{n}}{(2n)!}}=\frac{10}{(2n+1)(2n+2)}$$

Now, observe that $\frac{{a}_{n+1}}{{a}_{n}}<1$ for every $n\ge 1$. Thus, ${a}_{n+1}<{a}_{n}$ which means that $\{{a}_{n}\}$ is decreasing sequence.

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