limit of a n-square root and series of exponents

Question

Again, I'm having trouble with the infinite limits:

$$ (1) .... \lim_{n \to \infty} \sqrt[n]{ a^n+b^n } $$ with $a,b$ positive reals.

and to show if the following series is divergent or convergent

$$ (2) ......\sum_{n=1}^{\infty} \frac{5^{n}-2^{n}}{7^n-6^n} $$

To be honest, don't have any idea how to approach them, at least for the (2) I may use their exponential representations, as follows:

$$\sum_{n=1}^{\infty} \frac{e^{n \ln 5}-e^{n\ln 2}}{e^{n\ln 7}-6^{n \ln 6}} $$ and in that case the series will diverge. But for (1) don't know.

Thanks in advance!

Answer 1

For the first part: Without loss of generality let $a \ge b$. Then by binomial expansion

$$ \sqrt[n]{ a^n+b^n } = a(1 + (b/a)^n)^{1/n} = a\bigg(1 + \frac{(b/a)^n}{n} + ...\bigg) $$

Since $b/a <1$ each term after the first vanishes when $n \to \infty$. Hence

$$ \lim_{n \to \infty}\sqrt[n]{ a^n+b^n } = \max(a,b) $$

For the second part:

$$ \sum_{n=1}^{\infty} \frac{5^{n}-2^{n}}{7^n-6^n} = \sum_{n=1}^{5} \frac{5^{n}-2^{n}}{7^n-6^n} + \sum_{n=6}^{\infty} \frac{5^{n}-2^{n}}{7^n-6^n} < C + \sum_{n=6}^{\infty} \frac{5^{n}}{(6+1)^n-6^n} $$ where $C$ is the finite sum of the first five terms.

By binomial theorem, $$(6+1)^n - 6^n> (6^n + n6^{n-1} + \cdots + 1) - 6^n = n6^{n-1} + \cdots + 1 > 6^n \text{ for $n \ge 6$} $$

Hence $$ \sum_{n=1}^{\infty} \frac{5^{n}-2^{n}}{7^n-6^n} < C + \sum_{n=6}^{\infty} \frac{5^{n}} {6^n} < C + \sum_{n=1}^{\infty} \bigg(\frac{5} {6}\bigg)^n = C + 5 $$ which is convergent since $C$ is finite.

Answer 2

For the first part nothing to add, for the second we can observe that

$$\frac{\frac{5^{n}-2^{n}}{7^n-6^n}}{\frac1{{(7/5)}^{n}}}=\frac{7^{n}-(14/5)^{n}}{7^n-6^n}\to 1$$

then the given series converges by limit comparison test with $\sum_{n=1}^{\infty}\frac1{{(7/5)}^{n}}$.