TITLE: Non-induction proof for $\frac{1}{2\sqrt{2!}}+\frac{1}{3\sqrt[3]{3!}}+\cdots+\frac{1}{n\sqrt[n]{n!}}>\frac{n-1}{n+1}$
QUESTION [6 upvotes]: I know to prove it by induction but is this the only way ?
$$\frac{1}{2\sqrt{2!}}+\frac{1}{3\sqrt[3]{3!}}+\cdots+\frac{1}{n\sqrt[n]{n!}}>\frac{n-1}{n+1}, \space n\ge2$$

REPLY [4 votes]: It is hard to know what "without induction" may mean, since ultimately everything rests on induction. 
In this case, the Arithmetic Mean Geometric Mean Inequality will work. For 
$$\sqrt[k]{k!} \le \frac{1+2+\cdots +k}{k}.$$
So our expression is greater than a sum of terms of the shape $\frac{1}{1+2+\cdots+k}$. These can be written as $2\left(\frac{1}{k}-\frac{1}{k+1}\right)$. Now when we add there is telescoping.