Alternating sequence without L'Hopital rule

Question

I tried to solve this limit $$\lim_{n\to \infty} \ \sqrt[n]{\frac{3^n+(-1)^n}{n^2}}$$ by observing that $(-1)^n$ can have two values depending if $n$ is even or odd.

For $n$ is even, I have: $$\lim_{n\to \infty} \ \sqrt[2n]{\frac{3^{2n}+1^{2n}}{{4n}^2}}$$ than I tried extracting $3^n$ and got $$3\cdot\lim_{n\to \infty} \ \sqrt[2n]{\frac{1+(\frac{1}{3})^{2n}}{{4n}^2}}$$ After this, I did $$3 \cdot \ \lim_{n\to \infty} \exp\left[\ln{\frac{1+(\frac{1}{3})^{2n}}{{4n}^2}}\right]$$ and I don't know what to do next with this one. And for odd $n$ I also stop at this point.

Please help me using simple limit manipulation, it can be solved without using notations etc.

Answer 1

Hint: Squeeze your expression between two simpler ones.

$$ \frac{3^n - 1}{n^2} \leq \frac{3^n + (-1)^n}{n^2} \leq \frac{3^n + 1}{n^2} \text{.} $$

Answer 2

What you are left with in the end is $$\exp\left(\frac1{2n}\ln\left(\frac{1+\left(\frac13\right)^{2n}}{n^2}\right)\right)$$ You can decompose the logarithm in a difference, i.e. $$ \frac1{2n}\ln\left(\frac{1+\left(\frac13\right)^{2n}}{n^2}\right)=\frac1{2n}\ln\left(1+\left(\frac13\right)^{2n}\right)-\frac{\ln(n)}{n}$$ The first term trivially converges to zero, and it is known that the second also does.

So the whole exponential converges to $1$ and the limit you are looking for is $3$.

Answer 3

we can let: $$L(n)=\left(\frac{3^n+(-1)^n}{n^2}\right)^{1/n}$$ and because it is alternating we can say that: $$\left(\frac{3^n-1}{n^2}\right)^{1/n}\le L(n)\le\left(\frac{3^n+1}{n^2}\right)^{1/n}$$

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If we now focus on that upper limit: $$L_2=\left(\frac{3^n+1}{n^2}\right)^{1/n}=\exp\left(\frac{1}{n}\left[\ln(3^n+1)-2\ln(n)\right]\right)$$ now note that: $$\lim_{n\to\infty}(3^n+1)=\lim_{n\to\infty}3^n\tag{1}$$ and so we can say: $$\ln(3^n+1)-2\ln(n)=n\ln(3)-2\ln(n)$$ we can simplify our limit to: $$L_2=\exp\left(\frac{n\ln(3)-2\ln(n)}{n}\right)$$ $$\ln(L_2)=\ln(3)-\frac{2\ln(n)}{n}$$ which for $n\to\infty$ gives: $$\ln(L_2)=\ln(3)$$ and so $L_2=3$

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Now since our assumption of $(1)$ is also true for our lower limit, we can say: $$\lim_{n\to\infty}L(n)=3$$