# Thread: Does this limit proof work?

1. ## Does this limit proof work?

I have to prove that $a_{n}=\frac {1+2+3+...+n}{n^{2}}$ converges to 1/2.

Proof: We want to show that $a_{n}$ converges to 1/2. So we want to show that
$\left|\frac {1+2+3+...+n}{n^{2}}-\frac{1}{2}\right|<\epsilon$
Note that $a_{n}'=\frac {1+2+3+...+n}{n}>\frac {1+2+3+...+n}{n^{2}}=a_{n}$.
Thus if we prove the condition for $a_{n}'$ then we simultaneously prove it for $a_{n}$.
So $a_{n}'-\frac{1}{2}=\frac{n(n+1)}{2n}-\frac{1}{2}=\frac{n(n+1)-n}{2n}=\frac{n}{2}$.
So take $\frac{n}{2}<\epsilon \implies n<2\epsilon$.
So choose $n^{*}, and the condition holds for $a_{n}' \implies$ the condition holds for $a_{n}$ since $a_{n}. QED.

Is this a valid proof?

2. Why not use $\displaystyle\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$?

3. Originally Posted by paupsers
I have to prove that $a_{n}=\frac {1+2+3+...+n}{n^{2}}$ converges to 1/2.

Proof: We want to show that $a_{n}$ converges to 1/2. So we want to show that
$\left|\frac {1+2+3+...+n}{n^{2}}-\frac{1}{2}\right|<\epsilon$
Note that $a_{n}'=\frac {1+2+3+...+n}{n}>\frac {1+2+3+...+n}{n^{2}}=a_{n}$.
Thus if we prove the condition for $a_{n}'$ then we simultaneously prove it for $a_{n}$. This is not true.
So $a_{n}'-\frac{1}{2}=\frac{n(n+1)}{2n}-\frac{1}{2}=\frac{n(n+1)-n}{2n}=\frac{n}{2}$.
So take $\frac{n}{2}<\epsilon \implies n<2\epsilon$. You want n to be greater than some function of epsilon, not less than it.
So choose $n^{*}, and the condition holds for $a_{n}' \implies$ the condition holds for $a_{n}$ since $a_{n}. QED.

Is this a valid proof?
Proof:

$\left|\frac{\sum_{k=1}^n k}{n^2}-\frac{1}{2}\right|<\epsilon$

$\left|\frac{\frac{n(n+1)}{2}}{n^2}-\frac{1}{2}\right|<\epsilon$

$\left|\frac{n+1}{2n}-\frac{1}{2}\right|<\epsilon$

$\left|\frac{n+1-n}{2n}\right|<\epsilon$

$\left|\frac{1}{2n}\right|<\epsilon$

So for any $\epsilon>0$, let $n>\frac{1}{2\epsilon}$, and the condition will hold. $\square$