determining if sequence has upper bound 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?
 A: Notice the Riemann sum
$$\frac1n\sum_{k=1}^n \frac1{1+k/n} < \int_0^1\frac{dt}{1+t} = \log 2$$
A: 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.
A: 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 $\frac 12 < x_n < 1, n \in \mathbb{N}$.
A: 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 $$\lim_{n\rightarrow+\infty}\sum_{i=1}^n\frac{1}{n+i}=\ln2.$$
