Question

Q: show that if zn=(an+bn)1/n where 0<a<b, then lim(zn)=b.

Answer 1

Answer on Question #64536–Math–Real Analysis

Question

Show that

if $z_{n} = (a^{n} + b^{n})^{\frac{1}{n}}$

where $0 < a < b$ , then

$$\lim _ {n \rightarrow \infty} z _ {n} = b.$$

Solution

Note that

$$z _ {n} = \left(a ^ {n} + b ^ {n}\right) ^ {\frac {1}{n}} = \left(b ^ {n} \cdot \left(\frac {a ^ {n}}{b ^ {n}} + 1\right)\right) ^ {\frac {1}{n}} = \left(b ^ {n}\right) ^ {\frac {1}{n}} \cdot \left(\frac {a ^ {n}}{b ^ {n}} + 1\right) ^ {\frac {1}{n}} = b \cdot \left(\frac {a ^ {n}}{b ^ {n}} + 1\right) ^ {\frac {1}{n}} = b \cdot \left(\left(\frac {a}{b}\right) ^ {n} + 1\right) ^ {\frac {1}{n}}.$$

For $0 < a < b$ we get $0 < \frac{a}{b} < 1$ hence

$$\lim _ {n \rightarrow \infty} \left(\frac {a}{b}\right) ^ {n} = 0.$$ $$\lim _ {n \rightarrow \infty} z _ {n} = \lim _ {n \rightarrow \infty} (a ^ {n} + b ^ {n}) ^ {\frac {1}{n}} = \lim _ {n \rightarrow \infty} b \cdot \left(\left(\frac {a}{b}\right) ^ {n} + 1\right) ^ {\frac {1}{n}} = b \lim _ {n \rightarrow \infty} \left(\left(\frac {a}{b}\right) ^ {n} + 1\right) ^ {\frac {1}{n}} = b \cdot (0 + 1) ^ {0} = b \cdot 1 = b.$$

Answer: $b$ .

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