TITLE: Limit of $\sqrt[2n+1]{n^2+n}$
QUESTION [7 upvotes]: Prove $a_n=\sqrt[2n+1]{n^2+n}$ tends to $1$ as $n$ tends to infinity.
In the textbook, a hint was given: Let $a_n=1+h$, then $n^2+n=(1+h)^{2n+1}\gt \binom{2n+1}{3}(h^3)$
Then the consecutive steps are some algebraic manipulation, I managed to prove that $$\sqrt[3]{\frac{1}{n+1}+\frac{1}{n^2}}\gt \sqrt[2n+1]{n^2+n}-1\gt0$$ So $a_n$ tends to $1$ as $n$ gets sufficiently large.
But I don't understand why $a_n$ was set as $1+h$ at the first place, in retrospective, this does make calculations a lot convenient. And how did the author know when to use the binomial coefficient and compare the fourth term, it seems the author just plucked it out of thin air.

REPLY [2 votes]: Well here is an intuitive (but non-rigorous) argument $$(n^{2} + n)^{1/(2n + 1)} \approx (n^{2})^{1/(2n + 1)} = n^{2/(2n + 1)} \approx n^{1/n} \to 1$$ where the last result $n^{1/n} \to 1$ is pretty standard. The idea to express $a_{n} = 1 + h$ and then show $h \to 0$ is by R. Courant (if I recall correctly) and he used it to show that $n^{1/n} \to 1$ as $n \to \infty$. He put $n^{1/n} = 1 + h$ so that $$n = (1 + h)^{n} > \frac{n(n - 1)}{2}h^{2}$$ so that $0 < h^{2} < \frac{2}{n - 1} \to 0$. Thus $h \to 0$ and $n^{1/n} = 1 + h \to 1$.
Note that using $n^{1/n} \to 1$ we can analyze the behavior of sequence $a_{n} = \{P(n)\}^{1/Q(n)}$ where $P(n), Q(n)$ are polynomials and this sequence also has limit $1$ (only restriction is that coefficient of highest power of $n$ in $P(n)$ must be positive otherwise the sequence is not defined for all $n$ and $Q(n)$ is of positive degree). The fact follows more easily if we use logs. Clearly $$\log a_{n} = \frac{\log P(n)}{Q(n)}$$ and if $P(n) = a_{0}n^{k} + a_{1}n^{k - 1} + \cdots$ then we can use $$\frac{a_{0}}{2}n^{k} < P(n) < 2a_{0}n^{k}$$ for all $n$ after a certain point and then on taking logs we get $$\frac{\log(a_{0}/2)}{Q(n)} + k\cdot\frac{\log n}{Q(n)} < \frac{\log(P(n))}{Q(n)} < \frac{\log(2a_{0})}{Q(n)} + k\cdot\frac{\log n}{Q(n)}\tag{1}$$ Since $n^{1/n}$ tends to $1$ it follows that $(\log n)/n \to 0$ and therefore $(\log n)/Q(n) \to 0$. Applying squeeze theorem on $(1)$ we get $(\log P(n))/Q(n) \to 0$ and hence $a_{n} \to 1$. Just for completeness note that if coefficient of highest power of $n$ in $Q(n)$ is negative then the inequalities in equation $(1)$ are reversed but result remains the same.
In your case $P(n) = n^{2} + n, Q(n) = 2n + 1$. The technique of Courant can also be used on $a_{n} = \{P(n)\}^{1/Q(n)}$ directly if $Q(n)$ is a positive integer for all large $n$. As in case of $n^{1/n}$ we put $a_{n} = 1 + h$ so that $$P(n) = (1 + h)^{Q(n)} = (1 + h)^{q} > \binom{q}{r}h^{r}$$ We just need to choose $r$ such that degree of $$\binom{q}{r} = \binom{Q(n)}{r} = F(n)$$ as a polynomial in $n$ is higher than that of $P(n)$ and then we get $0 < h^{r} < P(n)/F(n)$ where $F$ has degree greater than $P$ and hence $h^{r}$ tends to $0$ and so $h \to 0$.
If $Q(n) < 0$ for all large $n$ then we use $b_{n} = 1/a_{n} = \{P(n)\}^{1/R(n)}$ where $R(n) = -Q(n)$ so that $R(n) > 0$ for all large $n$ and $b_{n} \to 1$ so that $a_{n} \to 1$.