Prove that a sequence is monotone increasing and bounded. Function is $$a_n = \frac{(n+1)^{\frac{1}{(n+1)}}}{n^{\frac{1}{n}}}$$
I've just managed to prove $$(n+1)^{\frac{1}{(n+1)}} < n^{\frac{1}{n}} \text{if n > 4} $$
I just need to prove that the above sequence is bounded above by 1 and is an increasing sequence. Finding limit is not a problem. I can use limit a/b to prove it is 1.
 A: $$\log a_n=\frac{\log(n+1)}{n+1}-\frac{\log n}n$$
Then $$\log a_{n+1}-\log a_n=\frac{\log(n+2)}{n+2}-\frac{2\log(n+1)}{n+1}+\frac{\log n}n$$
The second derivative of the function $f(x)=\frac{\log x}x$ is
$$\frac{2\log x -3}{x^3}$$
which is positive for $x>\exp(3/2)$. So $f$ is convex for these values. This shows that $\log a_{n+1}-\log a_n$ is positive for big enough $n$ (namely, for $n\ge 5$). The monotonicity of $\log$ implies that $a_n$ is increasing, too.
A: You already have that $a_n\leq 1$, hence you just need to prove that the sequence $\{a_n\}$ is eventually increasing, that is equivalent to proving that the sequence $\left\{n^{1/n}\right\}$ is eventually log-convex. So we just need to prove that the function $f(x)=\frac{\log x}{x}$ is convex over some interval of the form $(a,+\infty)$. Since:
$$ f''(x) = \frac{2\log x-3}{x^3} $$
with the choice $a=e^{3/2}$ everything works nicely.
A: You can show that n^(1/n) converges to 1. The proof could go as follows: Let a(n) be the sequence n^(1/n), and let b(n)= a(n) +1, so n^(1/n)= (1 + b(n)), which implies that n= (1 + b(n))^n, but (p + q)^n = 1 + np + (n(n-1)/2)*(p^2)q + ... (until nth term), so (p + q)^n > (n(n-1)/2)*(p^2)q, so n > (n(n-1)/2)*(b(n)^2), and hence, b(n) < sqrt(2)/sqrt(n-1), which converges to 0, and b(n)>0, so b(n) converges to 0, by Sandwich Lemma. Now, a(n) = b(n) + 1, so by Algebra of Convergent Sequences, a(n) converges to 1. Now, you know that the sequences in the denominator and the numerator of the given sequence c(n), converge to 1, and the denominator sequence n^(1/n) is never 0, so c(n), the given sequence, must converge to 1/1 = 1. 
