Question of monotone convergent sequence

Question

My Question is now

" Prove that $(2^n+3^n)^{1/n}$ is convergent. So I proved for two method by using monotone convergence theorem and squeeze theorem such as

1. monotone

   1) bounded below since 3 is lower bound

   2) decreasing (My real question!)

2. 3 < $(2^n+3^n)^{1/n}$ = $3(({2\over 3})^n+1)^{1/n}$ < $3({2\over 3}+1)^{1/n}$ so last sequence go to 3 "

In 1. - 2) , I can show that decreasing but it's very complex. My method is comparing $(2^n+3^n)^{n+1}$ and $(2^{n+1}+3^{n+1})^{n}$. Consequence : $(2^n+3^n)^{n+1}$ is lager than $(2^{n+1}+3^{n+1})^{n}$ so that $(2^n+3^n)^{1/n}$ is lager than $(2^{n+1}+3^{n+1})^{1\over{n+1}}$

Is there anyone who can prove decreasing part differently ?

Answer 1

How about

$$\left[\left(\frac23\right)^n+1\right]^{1/n}>\left[\left(\frac23\right)^{n+1}+1\right]^{1/n}>\left[\left(\frac23\right)^{n+1}+1\right]^{1/(n+1)}.$$

This seems more straightforward.

Answer 2

Let's call $a_n=(2^n+3^n)^{\frac 1n}$

$a_n>0$ thus $\ln(a_n)$ exists and

$\ln(a_n)=\frac 1n \times \ln(2^n+3^n)=\frac 1n \times \ln(3^n(1+(\frac 23) ^n)$

$\ln(a_n)=\frac 1n \times \ln(3^n) + \frac 1n \times \ln(1+(\frac 23) ^n)=ln(3)+\frac 1n \times \ln(1+u^n)$, $u=\frac 23 <1$

Thus $\lim_{n \to \infty}(2^n+3^n)^{\frac 1n}=3 $

Intuitively anyway $3^n$ will be always much greater than $2^n$ and the only significant term for $n$ high enough. You can discard the $2^n$ to find the limit. However you have to prove properly.

Answer 3

You can differentiate $(2^n + 3^n)^{\frac{1}{n}}$ and prove that it is a negative value over the desired domain (which I assume is $[1, \infty)$).

$\frac{d}{dn}((2^n + 3^n)^{\frac{1}{n}}) = (2^n + 3^n)^{\frac{1}{n}}\big(\frac{2^n \log(2) + 3^n \log(3)}{(2^n + 3^n)n} - \frac{\log(2^n + 3^n)}{n^2}\big)$

since we know $(2^n + 3^n)^{\frac{1}{n}}$ is positive, we only need to prove that

$\frac{2^n \log(2) + 3^n \log(3)}{(2^n + 3^n)n} \lt \frac{\log(2^n + 3^n)}{n^2} \rightarrow$

$n(2^n \log(2) + 3^n \log(3)) < (2^n + 3^n)\log(2^n + 3^n) \rightarrow$

$2^n \log(2^n) + 3^n \log(3^n) < 2^n\log(2^n + 3^n) + 3^n\log(2^n + 3^n)$

That should be enough to prove that it's decreasing.

Answer 4

Set $\frac{1}{n}=t$, log and rewrite the expression as $\log 3 + t \log (1 + x^{\frac{1}{t}})$ where $x= \frac{2}{3}<1$. Since $x^{\frac{1}{t}} \to_t 0$, then $\log (1 + x^{\frac{1}{t}}) \to_t 0$ too. Clearly the whole expression involving $\log$ tends to 0 from the above, hence the sequence is monotone decreasing.