does the sequence necessarily become decreasing??

**abhishekkgp** · December 28th 2011, 10:08 AM

Consider a sequence $(x_n)$ of real numbers with the following properties:
1) $x_n>0 \, \forall \, n$
2) $x_n<1 \, \forall \, n$
3) $x_{n+1}>x_n \, \forall \, n$

Define the sequence $(y_n)$ as $y_n=\frac{x_n}{n}$ .

Does $(y_n)$ necessarily become decreasing after a sufficiently large value of $n$ .

In other words, Does there exist a $N$ such that $y_{n+1} \leq y_n \, \forall \, n \, \geq N$ .

This is not a textbook question so i don't know the answer myself.
I can't think of an example of $(x_n)$ such that $y_{n+1} > y_n$ occurs frequently.

**Opalg** · December 28th 2011, 11:14 AM

Originally Posted by abhishekkgp

Consider a sequence $(x_n)$ of real numbers with the following properties:
1) $x_n>0 \, \forall \, n$
2) $x_n<1 \, \forall \, n$
3) $x_{n+1}>x_n \, \forall \, n$

Define the sequence $(y_n)$ as $y_n=\frac{x_n}{n}$ .

Does $(y_n)$ necessarily become decreasing after a sufficiently large value of $n$ .

In other words, Does there exist a $N$ such that $y_{n+1} \leq y_n \, \forall \, n \, \geq N$ .

This is not a textbook question so i don't know the answer myself.
I can't think of an example of $(x_n)$ such that $y_{n+1} > y_n$ occurs frequently.

The answer is No. In other words, such a sequence $(y_n)$ does not necessarily eventually decrease.

To start with, consider the sequence $(x_n)$ defined by $x_n = \tfrac k{k+1}$ for $k^2+1\leqslant n\leqslant (k+1)^2\ (k=0,1,2,\ldots).$ Then

$y_{k^2+1} = \frac{x_{k^2+1}}{k^2+1} = \frac k{(k+1)(k^2+1)},\qquad y_{k^2} = \frac{x_{k^2}}{k^2} = \frac {k-1}{k^3},$

and you can check by multiplying out the fractions that $y_{k^2+1} > y_{k^2}$ for all k. Thus there are infinitely many values of n for which $y_{n+1}>y_n.$

That example does not quite answer the original question, because the sequence $(x_n)$ is not strictly increasing (being constant throughout the interval $k^2+1\leqslant n\leqslant (k+1)^2$ ). However, in principle there is no difficulty in adjusting the sequence so as to make it strictly increasing. For each k, you can make $x_{k^2+1}$ very slightly smaller, without disturbing the property that $y_{k^2+1} > y_{k^2}$ . Then you can interpolate the values of $x_n$ linearly for $k^2+1\leqslant n\leqslant (k+1)^2$ so as to ensure that the sequence increases strictly.

**chisigma** · December 28th 2011, 06:47 PM

Originally Posted by abhishekkgp

Consider a sequence $(x_n)$ of real numbers with the following properties:
1) $x_n>0 \, \forall \, n$
2) $x_n<1 \, \forall \, n$
3) $x_{n+1}>x_n \, \forall \, n$

Define the sequence $(y_n)$ as $y_n=\frac{x_n}{n}$ .

Does $(y_n)$ necessarily become decreasing after a sufficiently large value of $n$ .

In other words, Does there exist a $N$ such that $y_{n+1} \leq y_n \, \forall \, n \, \geq N$ .

This is not a textbook question so i don't know the answer myself.
I can't think of an example of $(x_n)$ such that $y_{n+1} > y_n$ occurs frequently.

If we define $\delta_{x}(n)= x_{n+1}-x_{n}$ then it must be...

a) $\delta_{x}(n)>0\ \forall n$

b) $\lim_{n \rightarrow \infty} x_{n}= x_{1} + \sum_{k=1}^{\infty} \delta_{x}(k)= x_{0}\le 1$ (1)

Now, if $y_{n}=\frac{x_{n}}{n}$ we define $\delta_{y}(n)= y_{n+1}-y_{n}$ and in few steps we find that...

$\delta_{y}(n)= \frac{\delta_{x}(n)-y_{n}}{n+1}$ (2)

Because the series in (1) converges, it must be necesserly for some $\varepsilon>0$ and $N$ $\delta_{x}(n)<\frac{\varepsilon}{n}\ \forall n>N$ and that means that $\delta_{x}(n)<y_{n}\ \forall n>N$ so that $\forall n>N$ is $\delta_{y}(n)<0$ ...

Kind regards

$\chi$ $\sigma$

**abhishekkgp** · December 28th 2011, 11:54 PM

Originally Posted by Opalg

The answer is No. In other words, such a sequence $(y_n)$ does not necessarily eventually decrease.

To start with, consider the sequence $(x_n)$ defined by $x_n = \tfrac k{k+1}$ for $k^2+1\leqslant n\leqslant (k+1)^2\ (k=0,1,2,\ldots).$ Then

$y_{k^2+1} = \frac{x_{k^2+1}}{k^2+1} = \frac k{(k+1)(k^2+1)},\qquad y_{k^2} = \frac{x_{k^2}}{k^2} = \frac {k-1}{k^3},$

and you can check by multiplying out the fractions that $y_{k^2+1} > y_{k^2}$ for all k. Thus there are infinitely many values of n for which $y_{n+1}>y_n.$

That example does not quite answer the original question, because the sequence $(x_n)$ is not strictly increasing (being constant throughout the interval $k^2+1\leqslant n\leqslant (k+1)^2$ ). However, in principle there is no difficulty in adjusting the sequence so as to make it strictly increasing. For each k, you can make $x_{k^2+1}$ very slightly smaller, without disturbing the property that $y_{k^2+1} > y_{k^2}$ . Then you can interpolate the values of $x_n$ linearly for $k^2+1\leqslant n\leqslant (k+1)^2$ so as to ensure that the sequence increases strictly.

that was brilliant!! thanks. it solved my question.

**abhishekkgp** · December 29th 2011, 12:01 AM

Originally Posted by chisigma

If we define $\delta_{x}(n)= x_{n+1}-x_{n}$ then it must be...

a) $\delta_{x}(n)>0\ \forall n$

b) $\lim_{n \rightarrow \infty} x_{n}= x_{1} + \sum_{k=1}^{\infty} \delta_{x}(k)= x_{0}\le 1$ (1)

Now, if $y_{n}=\frac{x_{n}}{n}$ we define $\delta_{y}(n)= y_{n+1}-y_{n}$ and in few steps we find that...

$\delta_{y}(n)= \frac{\delta_{x}(n)-y_{n}}{n+1}$ (2)

Because the series in (1) converges, it must be necesserly for some $\varepsilon>0$ and $N$ $\delta_{x}(n)<\frac{\varepsilon}{n}\ \forall n>N$ and that means that $\delta_{x}(n)<y_{n}\ \forall n>N$ so that $\forall n>N$ is $\delta_{y}(n)<0$ ...

Kind regards

$\chi$ $\sigma$

thank you chisigma for your reply. It seems your answer to my question is "yes, the sequence necessarily becomes decreasing after a sufficiently large value of N." In that case i will take some time to review your solution. You may also want to read a very "close counterexample" provided by opalg.
thanks again.

**Opalg** · December 29th 2011, 12:08 AM

Originally Posted by chisigma

$x_{1} + \sum_{k=1}^{\infty} \delta_{x}(k)= x_{0}\le 1$ (1)

...

Because the series in (1) converges, it must be necessarly for some $\varepsilon>0$ and $N$ $\delta_{x}(n)<\frac{\varepsilon}{n}\ \forall n>N$

That is not true. Suppose for example that $a_n$ is always zero unless n is a perfect cube, with

$a_n = \begin{cases}1/\sqrt n = k^{-3/2} & \text{if }n=k^3, \\ 0 & \text{otherwise.} \end{cases}$

Then $\textstyle\sum a_n = \textstyle\sum k^{-3/2} < \infty$ . But if $n=k^3$ then $na_n = k^{3/2} \to\infty$ as $k\to\infty.$ Therefore there do not exist $\varepsilon>0$ and $N$ such that $a_n<\frac{\varepsilon}{n}$ for all $n>N.$

The point there, as in the OP's problem, is that a series or sequence can have bad behaviour sporadically while still being well-behaved as a whole.

**chisigma** · December 29th 2011, 09:17 PM

Originally Posted by chisigma

If we define $\delta_{x}(n)= x_{n+1}-x_{n}$ then it must be...

a) $\delta_{x}(n)>0\ \forall n$

b) $\lim_{n \rightarrow \infty} x_{n}= x_{1} + \sum_{k=1}^{\infty} \delta_{x}(k)= x_{0}\le 1$ (1)

Now, if $y_{n}=\frac{x_{n}}{n}$ we define $\delta_{y}(n)= y_{n+1}-y_{n}$ and in few steps we find that...

$\delta_{y}(n)= \frac{\delta_{x}(n)-y_{n}}{n+1}$ (2)

Because the series in (1) converges, it must be necesserly for some $\varepsilon>0$ and $N$ $\delta_{x}(n)<\frac{\varepsilon}{n}\ \forall n>N$ and that means that $\delta_{x}(n)<y_{n}\ \forall n>N$ so that $\forall n>N$ is $\delta_{y}(n)<0$ ...

May be it is necessary some more explanation of me so that I reported my post. The basic starting points are...

a) $\delta_{x}(n)>0\ \forall n$

b) $\lim_{n \rightarrow \infty} x_{n}= x_{1} + \sum_{k=1}^{\infty} \delta_{x}(k)= x_{0}\le 1$ (1)

... and the a) means that 'punctured infinite sums' like $\sum_{n} \delta_{n}$ where...

$\delta_{n}=\begin{cases} \frac{1}{\sqrt{n}} &\text{if}\ n=k^{3}\\ 0 &\text{otherwise}\end{cases}$ (2)

... aren't adequate counterexamples...

Kind regards

$\chi$ $\sigma$

**Opalg** · December 30th 2011, 12:28 AM

Originally Posted by chisigma

May be it is necessary some more explanation of me so that I reported my post. The basic starting points are...

a) $\delta_{x}(n)>0\ \forall n$

b) $\lim_{n \rightarrow \infty} x_{n}= x_{1} + \sum_{k=1}^{\infty} \delta_{x}(k)= x_{0}\le 1$ (1)

... and the a) means that 'punctured infinite sums' like $\sum_{n} \delta_{n}$ where...

$\delta_{n}=\begin{cases} \frac{1}{\sqrt{n}} &\text{if}\ n=k^{3}\\ 0 &\text{otherwise}\end{cases}$ (2)

... aren't adequate counterexamples...

If you want a more "adequate" counterexample, you could take

$\delta_{n}=\begin{cases} \frac{1}{\sqrt{n}} &\text{if}\ n=k^{3}, \\ \frac1{n^2}&\text{otherwise}.\end{cases}$

The fact is that the convergence of the series $\textstyle\sum\delta_n$ does not imply that there exist $\varepsilon>0$ and $N$ such that $\delta_n<\frac{\varepsilon}{n}$ for all $n>N$ .

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