Is the sequence $a_{n} = \frac {1}{2^2} + \frac{2}{3^2} + .... +\frac{n}{(n+1)^2}$ Cauchy?

Question

First:

I tried substituting natural numbers for $n$ to calculate the consecutive terms of the sequence and then see the difference between their values and I found that the difference is decreasing for large values of $n$ (not very large because I am calculating by my hand)so I concluded that it is a Cauchy sequence but unfortunately when I looked at the hint for solving this problem I found it to be: "Prove that $a_{2n} - a_{n} \geq n * \frac{2n}{(2n +1)^2} \geq \frac{2}{9}$" so I concluded that it is not Cauchy. So can anyone tell me please why testing by numbers sometimes lead to wrong values and when it is preferable to use this test?

Second:

If $a_{n} = \frac {1}{2^2} + \frac{2}{3^2} + .... +\frac{n}{(n+1)^2}$, Does $a_{2n} = \frac {1}{2^2} + \frac{2}{3^2} + .... +\frac{2n}{(2n+1)^2}$? because I am confused what it equals.

I hope my question fulfills the requirements of a good question, if no please let me know.

score 2 · Answer 1 · answered Sep 17 '18 at 00:49

2

Cauchy sequence requires $|a_m-a_n| \to 0$ when $n$ and $m$ both $\to \infty$. So the note that for $m=2n$ it does not $\to 0$ shows that the sequence is not Cauchy.

answered Sep 17 '18 at 00:49

herb steinberg

12,765

Is my formula to $a_{2n}$ written correctly? – Intuition Sep 17 '18 at 00:53
The formula for $a_{2n}$ is correct. It doesn't matter what it equals. The important point is that $|a_{2n}-a_n|$ does not $\to 0$. – herb steinberg Sep 17 '18 at 00:58
because I need to prove the formula in the given hint by induction ......this is why I am asking you if it is correct. – Intuition Sep 17 '18 at 01:03
Your first question is why use Cauchy? Your example illustrates why: $a_{2^k}={a_{2^k}}-a_{2^{k-1}}+a_{2^{k-1}}-a_{2^{k-2}}+.... \gt \frac{2}{9}k \to \infty$. – herb steinberg Sep 17 '18 at 01:10
My last comment might be useful if you want to use induction (on k, not on n). Is the induction to show $a_{2n}-a_n$ is bounded away from $0$? – herb steinberg Sep 17 '18 at 01:17
but why the $k$ is a power of the 2 in your before last comment? – Intuition Sep 17 '18 at 01:26
The basic idea is that your induction steps go from n to 2n to 4n, etc., because you are using $|a_{2n}-a_n|\gt \frac{2}{9}$ for each step. – herb steinberg Sep 17 '18 at 16:02

score 1 · Answer 2 · answered Sep 17 '18 at 01:08

1

Notice that

$$(n+1)^2 \le 4n^2 \iff n+1 \le 2n \iff n \ge 1$$

so $$a_n = \sum_{k=1}^n\frac{k}{(k+1)^2} \ge \sum_{k=1}^n\frac{k}{4k^2} = \frac14 \sum_{k=1}^n \frac1k$$

Therefore $(a_n)_n$ is unbounded so it cannot be Cauchy.

answered Sep 17 '18 at 01:08

mechanodroid

46,490

but I need to proof it by definition – Intuition Sep 17 '18 at 02:07
@hopefully Can you show by definition that Cauchy sequences are necessarily bounded? – mechanodroid Sep 17 '18 at 02:08
supposed to know but I think the above hint is given to precise $\epsilon$ , n and m – Intuition Sep 17 '18 at 02:10

score 1 · Accepted Answer · answered Sep 17 '18 at 02:23

"I tried substituting natural numbers for n to calculate the consecutive terms of the sequence and ...found that the difference is decreasing"

It doesn't matter if the difference between consecutive terms get smaller (and converge to zero). The terms $a_n$ and $a_{n+1}$ may get minisculely close together but in all the terms further down $a_{10^{\text{500 hundred quadrillion zillion oogleplex}}}$ can still be huge.

What does matter is that as $N$ gets large then the difference between any two terms (not just consecutive terms). To be Cauchy $a_n$ and $a_{10^{\text{500 hundred quadrillion zillion oogleplex}}}$ must be close together if $n, 10^{\text{500 hundred quadrillion zillion oogleplex}}$ are both greater than $N$.

So it doesn't matter of $a_{n+1} - a_n = \frac {n+1}{(n+2)^2} \to 0$. We need to have as $a_m - a_n = \frac {n+1}{(n+2)^2} + \frac {n+2}{(n+3)^2}+ \frac {n+3}{(n+4)^2}+ ....... + \frac {m}{(m+1)^2}\to 0$ as $n\to \infty$. As $a_m - a_n = \frac {n+1}{(n+2)^2} + \frac {n+2}{(n+3)^2}+ \frac {n+3}{(n+4)^2}+ ....... + \frac {m}{(m+1)^2}$ may have many, many,many terms this is a very tall order.

And it doesn't hold true.

$a_{2n} - a_n = \frac {n+1}{(n+2)^2} + \frac {n+2}{(n+3)^2} + ..... + \frac {2n}{(2n+1)^2} >$

$\frac {2n}{(2n+1)}^2 +\frac {2n}{(2n+1)}^2+.... + \frac {2n}{(2n+1)}^2 =$

$n*\frac {2n}{(2n+1)}^2 = \frac {2n^2}{4n^2 + 4n + 1} >$

$\frac {2n^2}{4n^2 + 4n^2 + n^2} =\frac {2n^2}{9n^2}= \frac 29$.

This means you will always be able to find to terms that are at least $\frac 29$ apart. No matter how large we take $N$ we can always find $n > N$ and $2n > N$ so that $a_{2n} - a_n > \frac 29$. So it is not true that the difference of any two terms goes to $0$.

So the sequence is not Cauchy.

I know that you increased the denominator so we decreased the fraction but for the numerator why must 2n be less than n +1 or n+2 or n+3 ..... — Intuition, Sep 17 '18 at 02:51
As $\frac {k}{(k+1)^2} > \frac {k+1}{(k+2)^2}$ for all $k \ge 1$. Although $k+1 > k$ then denominator simply grows at a faster rate to compensate for it. If you take the ratio $\frac {a_{k+1}}{a_k} = $$\frac {\frac {k+1}{(k+2)^2}}{\frac k{(k+1)^2}} =$$\frac{(k+1)(k+1)^2} {k(k+2)^2} = $$\frac {k^3 + 3k^2 + k + 1}{k^3 + 4k^2 + 4k} < 1$. So the $a_k$ are decreasing. — fleablood, Sep 17 '18 at 04:26

Is the sequence $a_{n} = \frac {1}{2^2} + \frac{2}{3^2} + .... +\frac{n}{(n+1)^2}$ Cauchy?

3 Answers3