Show that $(1+\frac 1n)^{n^2} \mathrm e^{-n}$ is not a zero sequence without logarithms

Question

How can I show that

$z_n = \left(1+\dfrac 1n \right)^{n^2} \mathrm e^{-n}$

is not a zero sequence? WolframAlpha says, it converges to $\frac{1}{\sqrt{e}}$ but how can I prove? I'd appreciate a solution without logarithms.

Answer 1

We will first argue that $z_n \ge e^{-1/2}$, by showing that that for $x \in [0,1], e^{x - x^2/2} \le 1 + x$.

TO do this, let $$ f(x) := e^{x - x^2/2} - 1 - x.$$ Note that, $$ f'(x) = (1-x) e^{x -x^2/2} - 1, \\ f''(x) = ((1-x)^2 -1)e^{x- x^2/2}$$

Note that the second derivative is non-positive for $x \in [0,1]$. Thus, the derivative is nonincreasing in ths domain. Since $f'(0) = 0,$ the derivative is non-positive in $[0,1]$ - i.e., $f$ is nonincreasing on $[0,1]$. Finally, since $f(0) = 0,$ this tells us that for $x \in [0,1], f(x) \le 0 \iff e^{x- x^2/2} \le 1 + x$.

Using this for $x = 1/n,$ we find that $ e^{1/n - 1/2n^2} \le 1 + 1/n,$ for $n \ge 1,$ and thus $$(1 + 1/n)^{n^2} e^{-n} \ge (e^{1/n - 1/2n^2})^{n^2} e^{-n} = e^{-1/2}.$$

We will now get an upper bound on the $z_n$. By truncating the series, we get that $e^{u} \ge 1 + u + u^2/2$ for $u \ge 0$. Now, let $x_n = \sqrt{1+2/n} - 1$. Note that $x_n + x_n^2/2 = 1/n$. So, we find that $(1+1/n) \le e^{x_n},$ and thus $$ (1 + 1/n)^{n^2} e^{-n} \le e^{n^2 x_n - n}. $$

Since $\exp(\cdot)$ is continuous, if we can argue that $ n^2 x_n - n \to -1/2,$ then we'd be done by the sandwich theorem. We can show this via a Taylor expansion: $\sqrt{1 + u} = 1 + u/2 - u^2/8 + O(u^3).$

Thus, $$ n^2 x_n - n = n^2 \left( 1 + \frac{(2/n)}{2} - \frac{(4/n^2)}{8} + O(n^{-3}) -1\right) - n = - \frac{1}{2} + O(n^{-1}),$$ and we're done.

Answer 2

Not a Zero-Sequence

In this answer, it is shown, using only an extension of Bernoulli's Inequality (which is proven there by induction), that $\left(1+\frac1n\right)^{n+1/2}$ is decreasing. Since the limit is $e$, this means $$ \left(1+\frac1n\right)^{n+1/2}\ge e\tag1 $$ Raising both sides to the $n-1/2$ power, we get $$ \left(1+\frac1n\right)^{n^2-1/4}\ge e^{n-1/2}\tag2 $$ Thus, $$ \left(1+\frac1n\right)^{n^2}e^{-n}\ge e^{-1/2}\left(1+\frac1n\right)^{1/4}\tag3 $$ Therefore, $$ \liminf_{n\to\infty}\left(1+\frac1n\right)^{n^2}e^{-n}\ge e^{-1/2}\tag4 $$ which is definitely non-zero.

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The Limit is $\boldsymbol{e^{-1/2}}$

To show the limit is $e^{-1/2}$ without logarithms requires a bit more work, but can be done using the same extension to Bernoulli's Inequality (case $m=2$ of Theorem $1$).

$$ \begin{align} \frac{\left(1+\frac1n\right)^{2n+1-2a}}{\left(1+\frac1{n-1}\right)^{2n-1-2a}} &=\frac{n+1}{n-1}\left(1-\frac1{n^2}\right)^{2n}\left(1+\frac1{n^2-1}\right)^{2a}\\ &\ge\scriptsize\frac{n+1}{n-1}\left(1-\frac2n+\frac{2n-1}{n^3}-\frac{(2n-1)(2n-2)}{3n^5}\right)\left(1+\frac1{n^2-1}\right)^{2a}\\ &=\left(1-\frac{n^2+2n-2}{3n^5}\right)\left(1+\frac1{n^2-1}\right)^{2a}\\ &\ge\left(1-\frac{n^2+2n-2}{3n^5}\right)\left(1+\frac{2a}{n^2}\right)\\ &=1+\frac{2a}{n^2}-\frac1{3n^3}-\frac2{3n^4}+\frac{2-2a}{3n^5}-\frac{4a}{3n^6}+\frac{4a}{3n^7}\tag5 \end{align} $$ Therefore, for any $a\gt0$, there is an $n_a$ so that for $n\ge n_a$, $$ \left(1+\frac1n\right)^{n+\frac12-a}\ge\left(1+\frac1{n-1}\right)^{n-\frac12-a}\tag6 $$ Since it is increasing to $e$, we have $$ \left(1+\frac1n\right)^{n+\frac12-a}\le e\tag7 $$ Raising both sides to the $n-\frac12+a$ power, we get $$ \left(1+\frac1n\right)^{n^2-\left(\frac12-a\right)^2}\le e^{n-\frac12+a}\tag8 $$ That is, for $n\ge n_a$, $$ \left(1+\frac1n\right)^{n^2}e^{-n}\le e^{-\frac12+a}\left(1+\frac1n\right)^{\left(\frac12-a\right)^2}\tag9 $$ Thus, for any $a\gt0$, $$ \limsup_{n\to\infty}\left(1+\frac1n\right)^{n^2}e^{-n}\le e^{-\frac12+a}\tag{10} $$ Therefore, $(4)$ and $(10)$ show that $$ \lim_{n\to\infty}\left(1+\frac1n\right)^{n^2}e^{-n}=e^{-1/2}\tag{11} $$

Answer 3

\begin{aligned} 0<\big(1+\frac1n\big)^{n^2}e^{-n}&=\exp\Big(n^2\big(\log\big(1+\frac1n\big)-n\Big)\\ &= \exp\Big(n^2\big(\frac1n-\frac{1}{2n^2}+\frac{1}{3n^3}-\ldots\big)-n\Big)\\ &=e^{-\frac12}\exp\Big(\frac{1}{3n}-\frac{1}{4n^2}+\ldots\Big)\xrightarrow{n\rightarrow\infty}e^{-\frac12} \end{aligned} To justify passing to the limit, notice that the series $f(x)=\sum_{n\geq1}(-1)^{n+1}\frac{x^n}{n+2}$ converges for $|x|<1$ and so, $\lim_{x\rightarrow0}f(x)=0$.