Sketch of a limit proof. Is this legal?

**CountingPenguins** · June 28th 2011, 01:04 AM

I'm going to start by saying I wish I knew all the tags to make this look nice.

The question is prove the lim (1/n - 1/(n+1)) = 0
My sketch of this is ((1(n+1)-n)/n(n+1))=0 which leads to ((n+1-n)/n(n+1)) which leads to (1/n(n+1)) and finally (1/((n^2)+n)) which I wholeheartedly expect converges to 0.

Now how do I make this formal enough to convince someone else, or am I off the mark?

**Sudharaka** · June 28th 2011, 01:20 AM

Originally Posted by CountingPenguins

I'm going to start by saying I wish I knew all the tags to make this look nice.

The question is prove the lim (1/n - 1/(n+1)) = 0
My sketch of this is ((1(n+1)-n)/n(n+1))=0 which leads to ((n+1-n)/n(n+1)) which leads to (1/n(n+1)) and finally (1/((n^2)+n)) which I wholeheartedly expect converges to 0.

Now how do I make this formal enough to convince someone else, or am I off the mark?

Dear CountingPenguins,

I hope you want to prove, $\lim_{n\rightarrow\infty}\left(\frac{1}{n}-\frac{1}{1+n}\right)=0.$ If that is the case the best method is to use the definition of limit of a function.

**Prove It** · June 28th 2011, 02:26 AM

Originally Posted by CountingPenguins

I'm going to start by saying I wish I knew all the tags to make this look nice.

The question is prove the lim (1/n - 1/(n+1)) = 0
My sketch of this is ((1(n+1)-n)/n(n+1))=0 which leads to ((n+1-n)/n(n+1)) which leads to (1/n(n+1)) and finally (1/((n^2)+n)) which I wholeheartedly expect converges to 0.

Now how do I make this formal enough to convince someone else, or am I off the mark?

The precise definition of a limit is that $\displaystyle \lim_{x \to \infty}f(x) = L$ if $\displaystyle x > N \implies |f(x) - L| < \epsilon$ .

So in this case, you are hoping to show $\displaystyle \lim_{n \to \infty}\left(\frac{1}{n} - \frac{1}{n + 1}\right) = 0$ , in other words, that $\displaystyle n > N \implies \left|\frac{1}{n} - \frac{1}{n + 1} - 0\right| < \epsilon$ .

$\displaystyle \begin{align*} \left|\frac{1}{n} - \frac{1}{n + 1}\right| &< \epsilon \\ \left|\frac{1}{n(n + 1)}\right| &< \epsilon \\ \frac{1}{|n(n + 1)|} &< \epsilon \\ |n(n + 1)| &> \frac{1}{\epsilon} \\ |n^2 + n| &> \frac{1}{\epsilon} \\ \left|n^2 + n + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \right| &> \frac{1}{\epsilon} \\ \left|\left(n + \frac{1}{2}\right)^2 - \frac{1}{4}\right| &> \frac{1}{\epsilon} \\ \left|n + \frac{1}{2}\right|^2 + \left|-\frac{1}{4}\right| &> \frac{1}{\epsilon} \textrm{ by the triangle inequality} \\ \left|n + \frac{1}{2}\right|^ 2 &> \frac{1}{\epsilon} - \frac{1}{4} \\ \left|n + \frac{1}{2}\right| &> \sqrt{\frac{1}{\epsilon} - \frac{1}{4}} \\ n + \frac{1}{2} &> \sqrt{\frac{1}{\epsilon} - \frac{1}{4}}\textrm{ since everything is positive} \\ n &> \sqrt{\frac{1}{\epsilon} - \frac{1}{4}} - \frac{1}{2}\end{align*}$

So by making $\displaystyle N = \sqrt{\frac{1}{\epsilon} - \frac{1}{4}} - \frac{1}{2}$ the proof will follow.

**Krizalid** · June 28th 2011, 04:26 PM

$\frac{1}{n(n+1)}<\frac{1}{{{n}^{2}}}<\epsilon ,$ so having $N=\left\lfloor \frac{1}{\sqrt{\epsilon }} \right\rfloor +1,$ also works for all $\epsilon>0.$

**CountingPenguins** · June 28th 2011, 08:27 PM

Thank you for your response. How did you know to add and then subtract (1/2)^2? Does that knowledge come with experience or is there a trick to it? Please let me know. Thanks.

**Also sprach Zarathustra** · June 28th 2011, 08:42 PM

So by making $\displaystyle N = \sqrt{\frac{1}{\epsilon} - \frac{1}{4}} - \frac{1}{2}$ the proof will follow

And natural $N$ is $\displaystyle \left \lfloor \sqrt{\frac{1}{\epsilon} - \frac{1}{4}} - \frac{1}{2} \right \rfloor +1$

**Prove It** · June 28th 2011, 09:18 PM

Originally Posted by CountingPenguins

Thank you for your response. How did you know to add and then subtract (1/2)^2? Does that knowledge come with experience or is there a trick to it? Please let me know. Thanks.

Completing the square. It's a powerful method when dealing with quadratic inequalities.

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