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