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

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