Found the following exercise in Bartle's Elements of Real Analysis in the section on combinations of sequences. Am unsure about my solution and would really appreciate it if someone could verify it.
If $ \; 0 \lt a \le b \; $ and $x_n = (a^n + b^n)^{\frac 1 n}, \; $ then $\lim (x_n) = b$
My Attempt:
$x_n = (a^n + b^n)^{\frac 1 n} = b\left[\left({\frac a b}\right)^n + 1\right]^{\frac 1 n} $ where $\frac a b \le 1$. Now let us define two more sequences in $\Bbb R, \;\; (y_n)$ and $(z_n)$ such that $y_n = b$ and $\; z_n = \left[\left({\frac a b}\right)^n + 1\right]^{\frac 1 n}$ for all $n \in \Bbb N$. And furthermore, since $\frac a b \le 1, $ let us equate the fraction to $\dfrac {1}{1 + t}$ for some $t \ge 0. $ Then:
$$|z_n - 1| = \left|{ \left[{\left({\dfrac {1}{1 + t}}\right)^n +1}\right]^{\frac 1 n} - 1}\right| = \left[{\left({\dfrac {1}{1 + t}}\right)^n +1}\right]^{\frac 1 n} - 1 $$
By Bernoulli's Inequality:
$$ |z_n - 1| \le \left[{\left({\dfrac {1}{1 + nt}}\right) +1}\right]^{\frac 1 n} - 1^{\frac 1 n} \le \dfrac {1}{(1 + nt)^{\frac 1 n}} \lt \dfrac{1}{(nt)^{\frac 1 n}}$$
Given any $\epsilon \gt 0$ there is a natural number $m$ such that $ m \gt \dfrac{1}{t\epsilon^m}$. Then;
$$n \ge m \implies \epsilon \gt \dfrac{1}{(tm)^{\frac 1 m}} \ge \dfrac{1}{(tn)^{\frac 1 n}} \gt |z_n - 1| \implies \lim (z_n) = 1$$
We know that $\lim (y_n) = b$. Therefore $(x_n) = (y_n.z_n)$ converges to $\lim(y_n).\lim(z_n) = b$
Q.E.D.
The following identities have also been used:
- $$\left|\sqrt[n]y-\sqrt[n]x\right|\le\sqrt[n]{|y-x|} \;\; \text{for every $x, y \gt 0$} $$
- $$n^n \lt (n + 1 )^{n+ 1} \;\; \forall n \in \Bbb N$$
Note that: $$b\leq(a^{n}+b^{n})^n=b((\frac{a}{b})^n+1)^\frac{1}{n}\leq b2^{\frac{1}{n}}$$