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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:

  • $$\left|\sqrt[n]y-\sqrt[n]x\right|\le\sqrt[n]{|y-x|} \;\; \text{for every $x, y \gt 0$} $$
  • $$n^n \lt (n + 1 )^{n+ 1} \;\; \forall n \in \Bbb N$$
Ishfaaq
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  • You have the inequality wrong; this is not true if $a = b$. – Ben Grossmann Mar 07 '14 at 02:00
  • @Omnomnomnom: You're right. And double checked the exercise. The condition is $0 \lt a \le b$. Any alternative suggestions. Just hints would be great if that is alright.. – Ishfaaq Mar 07 '14 at 02:12
  • @Omnomnomnom: Although I can solve for the case when $a = b$ separately. $(x_n)$ reduces to $(2)^{\frac 1 n}b$ – Ishfaaq Mar 07 '14 at 02:13
  • As the other answerers have pointed out, it seems I was wrong in my "observation". – Ben Grossmann Mar 07 '14 at 02:30
  • @Omnomnomnom: Oh.. I thought you meant my "proof" wouldn't work when $a = b$ which is actually true. Since in that case $t = 0$. Either way no harm done. – Ishfaaq Mar 07 '14 at 09:55

2 Answers2

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Note that: $$b\leq(a^{n}+b^{n})^n=b((\frac{a}{b})^n+1)^\frac{1}{n}\leq b2^{\frac{1}{n}}$$

Ethan Splaver
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Hint for an alternative approach: Since $0<a\leq b$ you can easyly show that $$ b\leq x_n\leq 2^\frac{1}{n}\cdot b $$ for each $n\in\mathbb N$. Then use the Sandwich principle.

sranthrop
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