Show that the following sequence $\{x_n\}$ is increasing and bounded above, and hence convergent.

Question

Show that the following sequence $\{x_n\}$ is increasing and bounded above, and hence convergent. $x_n = \frac{1}{ n + 1} +\frac{ 1}{ n + 2} + · · · +\frac{ 1}{ 2n }$.

I need to find how the sequence converges by showing how it increases and is bounded above.

Answer 1

Hints:

• Increasing: $$\begin{align} x_{n-1} &= \frac{1}{ n} +\frac{ 1}{ n + 1} + \dots +\frac{ 1}{ 2n-2 } \\ x_n &= \frac{1}{ n + 1} +\frac{ 1}{ n + 2} + \dots +\frac{ 1}{ 2n } \end{align}$$ so $x_n = x_{n-1} + \frac{ 1}{ 2n } + \frac{ 1}{ 2n -1 } - \frac{ 1}{ n }$. Can you show that $$\frac{ 1}{ 2n } + \frac{ 1}{ 2n -1 } - \frac{ 1}{ n } > 0$$ ?

• Bounded above: To get $x_n$, you sum $n$ terms, each of them being less than $\frac{1}{n}$.

Answer 2

$$ \begin{align} \sum_{k=n+1}^{2n}\frac1k &=\sum_{k=1}^{2n}\color{#C00000}{\frac1k}-\sum_{k=1}^n\color{#00A000}{\frac1k}\tag{1}\\ &=\sum_{k=1}^n\left(\color{#C00000}{\frac1{2k-1}+\frac1{2k}}-\color{#00A000}{\frac1k}\right)\tag{2}\\ &=\sum_{k=1}^n\left(\frac1{2k-1}-\frac1{2k}\right)\tag{3}\\ &=\sum_{k=1}^n\frac1{2k(2k-1)}\tag{4} \end{align} $$ Thus, as a sum of positive terms, the sum increases with $n$.

Comparing to a telescoping series, $$ \begin{align} \sum_{k=1}^n\frac1{2k(2k-1)} &=\frac12+\sum_{k=2}^n\frac1{2k(2k-1)}\tag{5}\\ &\le\frac12+\sum_{k=2}^n\frac1{2k(2k-2)}\tag{6}\\ &=\frac12+\frac14\sum_{k=2}^n\left(\frac1{k-1}-\frac1k\right)\tag{7}\\ &=\frac34\tag{8} \end{align} $$

Answer 3

Let $x_n$ be defined as above. We show $x_n$ is increasing by induction. For the base case $ x_1=1/2 < 1/3 + 1/4 = 7/12 = x_2 $ Suppose $x_n \geq x_{n-1}$. We have the following: $ x_{n+1} = x_n + \dfrac{1}{2n+1} + \dfrac{1}{2n+2} - \dfrac{1}{n+1} = x_n + \dfrac{1}{4n^2+6n+2} $ Thus $x_{n+1} > x_n$ and the sequence is increasing. To show the sequence is bounded we note $ x_n < 1/n + x_n = \sum_{k=n}^{2n} 1/k < \int_n^2n 1/x \, dx = \ln (2n) - \ln (n) = \ln (2) $ Thus the series is increasing and bounded above and therefore convergent.

Answer 4

$x_{n+1} - x_n=$
$= ( \frac{1}{ n + 2} +\frac{ 1}{ n + 3} + · · · +\frac{ 1}{ 2n }+\frac{ 1}{ 2n+1 }+\frac{ 1}{ 2n+2 })-( \frac{1}{ n + 1} +\frac{ 1}{ n + 2} + · · · +\frac{ 1}{ 2n })=$ $= \frac{ 1}{ 2n+1 }+\frac{ 1}{ 2n+2 }-\frac{1}{ n + 1} =$
$= \frac{ 1}{ 2n+1 }+\frac{ 1}{ 2n+2 }-\frac{2}{ 2n + 2} =$
$= \frac{ 1}{ 2n+1 }-\frac{ 1}{ 2n+2 } =$
$= \frac{ 1}{ 2n+1 }-\frac{ 1}{ 2n+2 } =$
$= \frac{ 2n+2-2n-1}{ (2n+1)(2n+2) } =$
$= \frac{ 2n+2-2n-1}{ (2n+1)(2n+2) } =$
$= \frac{ 1}{ (2n+1)(2n+2) } >0$
so

$x_{n+1} - x_n>0$
$x_{n+1} > x_n$

and the sequence is increasing

$\frac{ 1}{ n+1 } < \frac{1}{n}$
$\frac{ 1}{ n+2 } < \frac{1}{n}$
...
$\frac{ 1}{ 2n } < \frac{1}{n}$

we add and we have

$x_n <1$
so the sequence is bounded above