determining if sequence has upper bound

The largest term is the first, so an obvious upper bound is to set all terms equal to the first one and get $$ x_n < \frac{n}{n+1} <1. $$

You could also say that, since the last term is the smallest, one has $$ x_n > \frac{n}{2n} = \frac 12, $$

which means that $\frac 12 < x_n < 1, n \in \mathbb{N}$.

By C-S $$\sum_{i=1}^n\frac{1}{n+i}=1+\sum_{i=1}^n\left(\frac{1}{n+i}-\frac{1}{n}\right)=1-\frac{1}{n}\sum_{i=1}^n\frac{i}{n+i}=$$ $$=1-\frac{1}{n}\sum_{i=1}^n\frac{i^2}{ni+i^2}\leq1-\frac{1}{n}\frac{\left(\sum\limits_{i=1}^ni\right)^2}{\sum\limits_{i=1}^n(ni+i^2)}=1-\frac{1}{n}\frac{\frac{n^2(n+1)^2}{4}}{\frac{n^2(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}}=$$ $$=1-\frac{3(n+1)}{2(5n+1)}=\frac{7n-1}{10n+2}<\frac{7}{10}.$$ Actually, $$\ln2=0.6931...$$ Cauchy-Schwarz forever!

Actually, by calculus we can show that $$\lim_{n\rightarrow+\infty}\sum_{i=1}^n\frac{1}{n+i}=\ln2.$$

Notice the Riemann sum $$\frac1n\sum_{k=1}^n \frac1{1+k/n} < \int_0^1\frac{dt}{1+t} = \log 2$$