# Limit of $(1+\frac{1}{n^3})^{n^2}$

I have been trying to solve the limit of $$y_n = (1+\frac{1}{n^3})^{n^2}$$. Through graphical analysis, I have found that $$\lim_{n \to \infty} y_n = 1$$ Which can also be intuitively be understood as $$n^3 \geq n^2$$. Using Bernoulli's inequality, you can easily find that $$y_n \geq (1+\frac{n^2}{n^3}) \geq 1$$ I have also found that $$y_n - y_{n+1} \geq \left( 1+\frac{1}{(n+1)^3}\right)^{n^2} - \left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^2} = \left( 1+\frac{1}{(n+1)^3}\right)^{n^2} \left(1 - \left( 1+\frac{1}{(n+1)^3} \right) ^{2n+1} \right) = \left( 1+\frac{1}{(n+1)^3}\right)^{n^2} \left( \frac{1}{(n+1)^3} \right) \left( 1 + \left( 1 + \frac{1}{(n+1)^3} \right) + \cdots + \left( 1+\frac{1}{(n+1)^3}\right)^{2n} \right) \geq 1*0*2n\geq 0$$$$\implies yn \geq y_{n+1}$$ Thus, by using the monotone convergence theorem, we know $$y_n$$ converges and has a lower bound of $$1$$. I am however stuck at showing that $$\inf{\{y_n | n \geq 1\}} = 1$$, which would show that $$\lim_{n \to \infty} y_n = 1$$. Could I get a hint or a nudge in the right direction ?

PS: I cannot use exponential and logarithmic properties, nor l'hopital's rule, as we have not defined all these things in class

• I missed the part that you can't use limit defn of e. will delete my ans. Commented Mar 14, 2022 at 10:22
• Avoid the use of $*$ to denote multiplication. That's a common practice in programming but this is a math forum. Use a \times b to get $a \times b$ and a \cdot b to get $a \cdot b$. Commented Mar 14, 2022 at 11:01

You can use the (easily to verify) fact that $$\left(1+\frac 1m \right)^m < 3$$ for all $$m \in \mathbb{N}$$ as follows:

$$\begin{eqnarray*} \left(1+\frac 1{n^3} \right)^{n^2} -1 & = & \frac 1{n^3}\sum_{k=0}^{n^2-1}\left(1+\frac 1{n^3}\right)^k\\ & \leq & \frac 1{n^3}n^2\left(1+\frac 1{n^3}\right)^{n^2-1} \\ & \leq & \frac 1{n}\left(1+\frac 1{n^3}\right)^{n^3} \\ & \leq & \frac 3{n} \stackrel{n\to\infty}{\longrightarrow} 0\\ \end{eqnarray*}$$

You can use binomial expansion and force an upper bound that converges to 1. $$(1 + \frac1{n^3})^{n^2} = 1 + \sum_{k=1}^{n^2} {n^2 \choose k} \frac1{n^{3k}} \leq 1 + \sum_{k=1}^{n^2} \frac{n^{2k}}{k!} \frac1{n^{3k}}$$

• An upper bound is definitely not sufficient to prove the convergence to $1$. Even if it is obvious, you should add the lower-bound part. Commented Mar 14, 2022 at 10:51
• Of course, one needs both for a complete proof. I didn't include it in the answer because the OP already had the lower bound. Commented Mar 14, 2022 at 10:54
• Oh ok, I did not see that the OP explicitely said something about the lower bound. Everything is fine, then ! Commented Mar 14, 2022 at 10:56

We have

$$y^n =(1+\frac{1}{n^3})^{n^3} \to e$$

as $$n \to \infty.$$ Hence there is $$N$$ such that

$$2 \le y^n \le 3$$

for $$n>N.$$

Can you proceed ?

• I wrote this too but it seems OP has not defined e as such so far Commented Mar 14, 2022 at 10:32