2

$$\lim_{n\to\infty}\left(\frac{2n+3}{2n-1}\right)^{n+1}$$

I have this, but I don't know how to start it. I tried rationalizing it, or factorizing it but it didn't work. Is there anyone who can give me any tips on how to do it?

grg
  • 1,017
John
  • 53

2 Answers2

3

$$\ \lim_{n\to\infty}\bigg(\frac{2n+3}{2n-1}\bigg)^{n+1}=\lim_{n\to\infty}\bigg(\frac{2n-1+4}{2n-1}\bigg)^{n+1}=\lim_{n\to\infty}\bigg(1+\frac{4}{2n-1}\bigg)^{n+1}=$$ $$\ =\lim_{n\to\infty}\bigg[\bigg(1+\frac{4}{2n-1}\bigg)^{2n+2}\bigg]^{\frac{1}{2}}=\lim_{n\to\infty}\bigg[\bigg(1+\frac{4}{2n-1}\bigg)^{2n-1}\cdot\bigg(1+\frac{4}{2n-1}\bigg)^{3}\bigg]^{\frac{1}{2}}=$$ $$=(e^4)^{\frac{1}{2}}\cdot(1)^{\frac{3}{2}}{}=e^2$$

Since $\ \forall\alpha,\beta \in \mathbb R, \lim_{a_n\to\infty}(1+\frac{\alpha}{a_n})^{\beta}=1$

Mosk
  • 1,630
  • This is very helpful till here. I remember the professor rolling something with -1, but i can't remember what did he do.

    I know i must bring this on the "E" form (1 + 1/an )^an but i can not see the next step here.

     – John Feb 19 '15 at 12:16
  • 1
    Oh so you added the 1/2 and then changed the n+1 to 2n+2. Then you divided it into two formulas since 2n-1+3 = 2n+2.

    Then the first formula fits for the "e", and the second based on the rule you posted turns out to be 1. The formula for the "e" is (1+1/an)^an but since you had (1+4/an)^an, you didnt write e^1 instead u wrote e^4.

    And then just e^4/2 = e^2.

    Thank you, i think i understand what is going on here, and how you should do it. I will look up internet for different examples to try and see if i can get them done now.

    Thank you a lot

     – John Feb 19 '15 at 12:37
2

Let $m=\frac{2n-1}{4}$.

$$\begin{align}\lim_{n\to\infty}\left(\frac{2n+3}{2n-1}\right)^{n+1}&=\lim_{n\to\infty}\left(1+\frac{4}{2n-1}\right)^{n+1}=\\=\lim_{n\to\infty}\left(1+\frac{1}{m}\right)^{2m+1.5}&=\lim_{m\to\infty}\left(1+\frac{1}{m}\right)^{2m+1.5}\end{align}$$

, because $n\to\infty\iff m\to\infty$.

$$\lim_{m\to\infty}\left(1+\frac{1}{m}\right)^{2m+1.5}=\lim_{m\to\infty} \left(\left(1+\frac{1}{m}\right)^m\right)^2\left(1+\frac{1}{m}\right)^{1.5}=e\cdot e\cdot \lim_{m\to\infty}\left(1+\frac{1}{m}\right)^{1.5}=e^2$$

user26486
  • 11,331
  • why do you have 2n-1/4, and based on Mosks tip i have 4/2n-1 ? – John Feb 19 '15 at 12:16
  • @John I let $m=\frac{2n-1}{4}$. As $n\to\infty$, $m$ also goes to $\infty$. $1+\frac{4}{2n-1}=1+\frac{1}{m}$ and $n+1=2m+1.5$ holds too. So we get that $\lim_{n\to\infty}\left(1+\frac{4}{2n-1}\right)^{n+1}=\lim_{m\to\infty}\left(1+ \frac{1}{m}\right)^{2m+1.5}$. – user26486 Feb 19 '15 at 12:19