Show that the sequence is monotonic

Its rather easy to show that $a_n=\frac{n^{1/n}}{n}$ ist monotonic (which means $a_{n+1}<a_n$ for each $n$) using derivations. But how can I do it without them? Thanks.

Solution 1:

Assume $n$ is positive, we show that the sequence is strictly decreasing (i.e. $a_n > a_{n+1}$).

Take $-n(n+1)$ power to both sides, you get the required $$n^{(n-1)(n+1)} = n^{n^2-1} < (n+1)^{n^2}.$$

Solution 2:

Simply for every $n>0$:
$\frac {a_(n+1)}{a_n}<(n+1)^(\frac {1}{n+1}-\frac {1}{n})<1$ , Thus $a_n $ is strictly decreasing .

Solution 3:

if by monotonic you mean sequences which constantly increase or constantly decrease, then take the logarithmic difference of two consecutive elements and show that it is always positive or negative (in your case positive). So if we set $${a_n} = {n^{\frac{1}{n} - 1}},$$ then $$\log {a_{n + 1}} - \log {a_n} = \left( {\frac{1}{{n + 1}} - 1} \right)\log \left( {n + 1} \right) - \left( {\frac{1}{n} - 1} \right)\log \left( n \right)$$

do the algebra and show the difference always positive. The use of algorithm function since it preserves monotonicity