Compute $\lim_{n\to\infty} (1+n+n^2)^{\frac{1}{n}}$.

Question

Compute $\lim_{n\to\infty} (1+n+n^2)^{\frac{1}{n}}$.

Specifically how to apply the binomial theorem to the function that we are taking the limit of? My calculus professor says we need to make use of the binomial theorem.

Answer 1

The following is a variation on the standard proof that $n^{1/n} \to 1$ as $n \to \infty$. And you can see that the usual binomial theorem for positive integral index is used here in a crucial manner.

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You can observe that $1 + n + n^{2} > 1$ and hence $(1 + n + n^{2})^{1/n} > 1$ and therefore if we set $(1 + n + n^{2})^{1/n} = 1 + x_{n}$ then $x_{n} > 0$. So we have $$1 + n + n^{2} = (1 + x_{n})^{n} = \sum_{r = 0}^{n}\binom{n}{r}x_{n}^{r}$$ Since $x_{n} > 0$ it is obvious that if we just take any specific term in the binomial expansion on RHS then this single term is less than the whole binomial sum and hence less than $1 + n + n^{2}$. Taking the term corresponding to $r = 3$ we get $$\binom{n}{3}x_{n}^{3} < 1 + n + n^{2}$$ for all $n \geq 3$ or $$0 < x_{n}^{3} < \frac{6(1 + n + n^{2})}{n(n - 1)(n - 2)}$$ and by Squeeze Theorem we get $$\lim_{n \to \infty}x_{n}^{3} = 0$$ and using continuity of $f(x) = x^{1/3}$ we see that $$\lim_{n \to \infty}x_{n} = 0$$ and hence $$\lim_{n \to \infty}(1 + n + n^{2})^{1/n} = \lim_{n \to \infty}(1 + x_{n}) = 1$$

Answer 2

The binomial theorem seems to me less important than this well-known limit:

$n^{\frac{1}{n}}\rightarrow 1$

Do you see how to proceed from here?

Answer 3

hint: $1 < 1+\dfrac{1}{n} + \dfrac{1}{n^2} < 3$

Answer 4

Consider $$A_n= (1+n+n^2)^{\frac{1}{n}}\implies \log(A_n)=\frac 1n \log(1+n+n^2)=\frac 1n \left(\log(n^2)+\log(1+\frac 1 n+\frac 1 {n^2})\right)$$ Now, using for small values of $x$ $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$ replace $x$ by $\frac 1 n+\frac 1 {n^2}$ (here will appear the binomial expansion).

This gives $$\log(1+\frac 1 n+\frac 1 {n^2})=\frac{1}{n}+\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ So $$\log(A_n)=\frac 1n \left(2\log(n)+ \frac{1}{n}+\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)\right)=2\frac{\log(n)}{n}+\frac 1 {n^2}+O\left(\frac{1}{n^3}\right)$$

I am sure that you can take it from here.