How to prove by induction $(n+1)^{1/n}$ is bounded above? Was wondering if anyone could help with the problem above. It seems like the sequence is bounded by $2$ and I proceeded with the induction steps to show all elements are at most $2$.
Base case $n=1$ holds true because $2$ is at most $2$.
Induction assumption for some value $k$, from which I get $(k+2)^{1/(k+1)}\le 2$
But how do I proceed from here? Did I get here correctly? Do I use logarithms to proceed?
Thanks! (Sorry for the unformatted post).
Edit: sequence begins with $n=1,2,3\ldots$
 A: An idea:
Begin with $\;n<2^n\;$ forall $\;n\in\Bbb N$ :
$$\;\forall\,n\in\Bbb N\;,\;\;\;n<2^n\implies n^{1/n}<(2^n)^{1/n}=2\iff \sqrt[n]n<2\,,\,\,\forall\,n\in\Bbb N\;$$
If you want, you can carry on induction to prove $\;n<2^n\;$ for all $\;n=1,2,3,...\;$ N;$
A: Consider the sequence $a_n = (n + 1)^{1/n}$ defined for all integers $n \geq 1.$ Using logarithmic differentiation on the continuous function $f(x) = (x + 1)^{1/x}$ defined for all real numbers $x \geq 1,$ we can prove that $f(x)$ is decreasing, hence $f(x)$ attains an absolute maximum. Prove that $f(x)$ takes its maximum value of $2$ at $x = 1,$ hence the sequence $a_n$ is bounded above by $2.$
Once you believe this, you can proceed by induction to prove that $a_n \leq 2$ for all integers $n \geq 1,$ if you wish. For $n = 1,$ the claim holds, hence we may assume inductively that $a_n \leq 2.$ Observe that $$a_{n + 1} \leq 2 \iff \ln a_{n + 1} = \ln (n + 2)^{1 / (n + 1)} = \frac{\ln(n + 2)}{n + 1} \leq \ln 2$$ because $\ln x$ and $e^x$ are increasing functions. Consequently, it suffices to show that $\ln(n + 1) \leq (\ln 2)n$ for all integers $n \geq 0$; this can be accomplished via induction.
Ultimately, this shows that there are better ways to prove that $(n + 1)^{1/n}$ is bounded above by $2.$
