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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.

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    $\begingroup$ Change the index of the sum such that you had $i$ instead of $n+i$ in the denominator. $\endgroup$ – Masacroso Nov 1 '16 at 5:54
  • $\begingroup$ You should get $x_n - x_{n+1} = \frac{1}{4n^2+6n+2}$. Just expand and change indices. $\endgroup$ – астон вілла олоф мэллбэрг Nov 1 '16 at 5:57
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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 $$\lim_{n\rightarrow\infty}x_{n}=\color{red}{1-\log\left(2\right)}.$$

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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 $.

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