$d_n=(1+(2/n))^n$ converge or diverge and find the limit?

Question

I know the answer is $e^2$ and I'd like to use L'Hopital's rule because this is an indeterminate form. Can someone explain how to get there?

$$d_n=\left(1+\frac{2}{n}\right)^n$$

Answer 1

If you know that the sequence converges, then note that since the subsequence $d_{2n} = (1 + 1/n)^{2n} = [(1 + 1/n)^n]^2$ converges to $e^2$, so does $d_n$.

Answer 2

Using the well known limit $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$$

we can get $$\begin{align} \lim_{n\to\infty}\left(1+\frac2n\right)^n & =\lim_{n\to\infty}\left(1+\frac1{n/2}\right)^n\\ & = \lim_{m\to\infty}\left(1+\frac1m\right)^{2m}\quad m=n/2\\ & = \lim_{m\to\infty}\left[\left(1+\frac1m\right)^m\right]^2\\ & = e^2 \end{align}$$

Answer 3

A term of the form $f(n)^{g(n)}$ can usually be converted to a L'Hopital's rule form by taking the log of both sides.

$$\ln d_n = \ln \left(\left(1+\frac 2n\right)^n\right) = n \ln \left(1+\frac 2n\right) = \frac{\ln\left(1+\frac 2n\right)}{\frac 1n}$$

From here you can probably show that

$$\lim_{n\rightarrow\infty} \ln d_n = 2.$$

Then, since $\ln$ is continuous,

$$\lim_{n\rightarrow\infty} \ln d_n = \ln \lim_{n\rightarrow\infty} d_n = 2,$$

and you can solve to get

$$\lim_{n\rightarrow\infty} d_n = e^2.$$

Answer 4

Apply $\ln$ to get

$$n \cdot \ln (1+2/n) =2\frac{\ln (1+2/n) -\ln 1}{2/n} \to 2\ln'(1) = 2.$$

No need for L'Hopital, just the definition of the derivative. Exponentiating back gives $e^2$ for the limit.