I am somewhat stuck in my calculations when determining if sequence has an upper bound.
The sequence $$x_n = \frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n-1}+\frac{1}{2n}$$ Is equal to $$\frac{1}{n}(\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+..+\frac{1}{1+\frac{n}{n}})$$
And so I notice that all the denominators are greater than 1, which means that all terms in the parentheses are less than 1.
But how can I determine further if there is an 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 $$\frac1n\sum_{k=1}^n \frac1{1+k/n} < \int_0^1\frac{dt}{1+t} = \log 2$$
hint
For each $n\ne 0$,
$$\frac{1}{n+1}\le \frac{1}{n}$$ $$\frac{1}{n+2}\le \frac{1}{n}$$ ...
$$\frac{1}{2n}\le \frac 1n$$
You can finish.
