0

The procedure I am about to write is wrong, and I know why it is wrong yet it was the only one I could think of, so I will put it up anyway $$\lim_{n\to \infty} \frac{e^n}{\left(1+\frac 1n\right)^{n^2}}$$

For the denominator $$\begin{aligned}\lim_{n\to \infty} \left(1+\frac 1n\right)^{n^2} &=e^{\lim_{n\to \infty} (1+\frac 1n-1)n^2} \\ &=e^{\lim_{n\to \infty} n}\end{aligned}$$

And the same for the numerator, so their division should give $1$

As I said, I know this is wrong. The correct answer is $\sqrt e$. What is the right process?

V.G
  • 4,196
Aditya
  • 6,191

2 Answers2

4

Note that

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

is an indeterminate form $1^\infty$ therefore we can't conclude that it is equal to one.

By $x=\frac1n \to 0^+$ we obtain

$$\large{\frac{e^n}{(1+\frac 1n)^{n^2}}=e^{\log\left(\frac{e^n}{(1+\frac 1n)^{n^2}}\right)}=e^{\frac{\frac1n-\log \left(1+\frac 1n\right) }{\frac1{n^2}}}}=e^{\frac{x-\log (1+x)}{x^2}}$$

then we can proceed by l'Hopital or Taylor's series to find

$$\frac{x-\log (1+x)}{x^2} \to \frac12$$

Assuming the limit exists, we can also use the result shown here:

user
  • 154,566
  • How did you write $\log (\frac{e^n}{(1+\frac 1n)n^2} $ equal to whatever you got? I solved it to obtain $n-\log (1+\frac 1n)-2\log n$ – Aditya Jul 28 '20 at 05:48
  • He just take the power $ n^2 $ out and then express it as logarithmic rule. – A learner Jul 28 '20 at 05:52
  • 1
    We have used $$\log\left(\frac{e^n}{(1+\frac 1n)^{n^2}}\right)=\log (e^n)-\log\left(1+\frac 1n\right)^{n^2}=n-n^2\log\left(1+\frac 1n\right)=n^2\left(\frac1n-\log\left(1+\frac 1n\right)\right)$$ – user Jul 28 '20 at 05:52
2

The numerator and the denominator both tend to $\infty$. You cannot conclude that the ratio tends to $1$. Instead,

$$e^{n-n^{2}\log (1+\frac 1 n)}=e^{n-n^{2}(\frac 1n -\frac 1{2n^{2}}+\frac 1 {3n^{3}}+\dotsb)} =e^{\frac 1 2 - \frac 1 {3n}+\dotsb}\to e^{\frac 1 2 }.$$

[For $n \geq 2$, the series $-\frac 1 {3n}+\frac1 {4n^{2}}+\dotsb$ is dominated by $ \frac 1 {(3)(2)} +\frac 1 {4(2^{2})}+\frac 1 {52^{3}}+\dotsb$, which is a convergent series. We can use DCT (or uniform convergence of the series) to conclude that $-\frac 1 {3n}+\frac1 {4n^{2}}+\dotsb$ converges to 0.]

SvanN
  • 2,307
Kavi Rama Murthy
  • 311,013