Prove that $x_n = 1 - \sum\limits^{n}_{i=1} \frac{1}{n+i}$ converges 
Let $(x_n)^{\infty}_{n=1}$ be the sequence $\displaystyle{x_n = 1 - \sum^{n}_{i=1} \frac{1}{n+i}}$. Prove that $(x_n)^{\infty}_{n=1}$ converges.

I know the sequence is decreasing and bounded below, hence will converge (Monotone Convergence Theorem), but I am having trouble actually proving that the sequence is decreasing.
 A: If
$x_n = 1 - \sum^{n}_{i=1} \frac{1}{n+i}
$,
then
$\begin{array}\\
x_{n+1}-x_n
&=\left(1 - \sum^{n+1}_{i=1} \frac{1}{n+1+i}\right)-\left(1 - \sum^{n}_{i=1} \frac{1}{n+i}\right)\\
&=\sum^{n}_{i=1} \frac{1}{n+i}-\sum^{n+1}_{i=1} \frac{1}{n+1+i}\\
&=\sum^{n}_{i=1} \frac{1}{n+i}-\sum^{n+2}_{i=2} \frac{1}{n+i}\\
&=\frac1{n+1}-\frac1{2n+1}-\frac1{2n+2}\\
&=\frac1{2n+2}-\frac1{2n+1}\\
&=\frac{(2n+1)-(2n+2)}{(2n+2)(2n+1)}\\
&=\frac{-1}{(2n+2)(2n+1)}\\
&< 0\\
\end{array}
$
Therefore
$x_{n+1}<x_n$.
Convergence is also easily proved
by comparison with
$\frac1{n^2}$
or,
more easily,
$\frac1{n(n+1)}$.
Explicitly,
$x_{n+1}-x_n
=\frac1{2n+2}-\frac1{2n+1}
>\frac1{2n+2}-\frac1{2n}
=\frac12(\frac1{n+1}-\frac1{n})
$
and this telescopes
so that,
if $m > n$,
$\begin{array}\\
x_m-x_n
&=\sum_{k=n}^{m-1} (x_{k+1}-x_k)\\
&>\sum_{k=n}^{m-1} (\frac12(\frac1{k+1}-\frac1{k}))\\
&=\frac12(\frac1{m}-\frac1{n})\\
&>-\frac1{2n}\\
\end{array}
$
so that,
if $m > n$,
$-\frac1{2n}
<x_m-x_n
< 0
$.
A: Note that we can prove the convergence also using the Riemann' sum. We have $$\sum_{i=1}^{n}\frac{1}{n+i}=\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+i/n}\underset{n\rightarrow\infty}{\rightarrow}\int_{0}^{1}\frac{1}{1+x}dx=\log\left(2\right)
 $$ hence $$\lim_{n\rightarrow\infty}x_{n}=\color{red}{1-\log\left(2\right)}.$$
