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I have the limit of an alternating sequence:

$$\lim_{n\rightarrow \infty} (-1)^n \frac{3^{2n+1}n^{2n}}{(4n-1)^{2n}}$$

I understand that if the limit of the absolute value of the sequence is $0$, then the sequence is convergent.

$$\lim_{n\rightarrow \infty}\frac{3^{2n+1}n^{2n}}{(4n-1)^{2n}}$$

But I don't know how to approach this limits, or what techniques to use.

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  • $\begingroup$ Sequence or series. If the limit of absolute value of the sequence is $0$ then the limit of the sequence is $0$. $\endgroup$
    – fleablood
    Commented Apr 8, 2021 at 16:43
  • $\begingroup$ We know that for any real number that $-|x| \le x \le |x|$. So for any sequence $-|a_n| \le a_n \le |a_n|$. So if we know $|a_n| \to 0$ then we know $-|a_n| \to 0$ and by the squeeze theorem $a_n \to 0$ $\endgroup$
    – fleablood
    Commented Apr 8, 2021 at 16:48
  • $\begingroup$ Wait... Is your question how do we know $\lim_{n\rightarrow \infty}\frac{3^{2n+1}n^{2n}}{(4n-1)^{2n}}=0 $? Or is it about if we know that $\lim_{n\rightarrow \infty}\frac{3^{2n+1}n^{2n}}{(4n-1)^{2n}} = 0$ how do we know what $\lim_{n\rightarrow \infty} (-1)^n \frac{3^{2n+1}n^{2n}}{(4n-1)^{2n}}$ is? or is it what is $\sum_{n=0}^{\infty} (-1)^n \frac{3^{2n+1}n^{2n}}{(4n-1)^{2n}}$? Or is it something else entirely. $\endgroup$
    – fleablood
    Commented Apr 8, 2021 at 16:51

1 Answer 1

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Note that we can write

$$\frac{3^{2n+1}n^{2n}}{(4n-1)^{2n}}=3\left(\frac{3n}{4n-1}\right)^{2n}=3\left(\frac{3/4}{1-\frac1{4n}}\right)^{2n}$$

Can you conclude now?

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  • $\begingroup$ Yes this helps out a lot, thank you! $\endgroup$
    – chubbygrass
    Commented Apr 8, 2021 at 17:03
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$
    – Mark Viola
    Commented Apr 8, 2021 at 17:03

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