## Thomas' Calculus 13th Edition

converges to $1$
As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists. Consider $a_n= \sqrt[n] {2n+1}$ or, $a_n= (2n+1)^{1/n}$ Apply limits to both sides. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty} (2n+1)^{1/n}$ Let us consider $y=\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty} (2n+1)^{1/n}$ Use logarithmic rule: $\ln a^n=n \ln a$ $\ln y=\lim\limits_{n \to \infty} \ln (2n+1)^{1/n}$ $\ln y=\lim\limits_{n \to \infty} \dfrac{\ln (2n+1)}{n}$ Since, we can see that the limit has the form of $\frac{\infty}{\infty}$, so take the help of L-Hospital's rule. $\ln y=\lim\limits_{n \to \infty} (\dfrac{2}{2n+1})(\dfrac{1}{1})$ $\ln y=0$ $e^{\ln y}=e^0 \implies y=1$ Therefore, the sequence converges to $1$.