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