Please evaluate my sketch of below proof on scale 1-5 in clarity, conciseness, and validity.

Show $\lim_{n\rightarrow\infty}(\frac{1}{n^2}-\frac{1}{n})=0$, $n\in\mathbb{R}$ using the epsilon-delta limit definition.
Sketch of the Proof:

For simplicity, assume ($\mathbb{R},d$) with $d$, the usual Euclidean metric on $\mathbb{R}$.
Use triangle inequality to break down the inside of limit expression as $\mid\frac{1}{n^2}-0\mid$+$\mid\frac{1}{n}-0\mid$.
Denote $(x_n)=\frac{1}{n^2}$,$(y_n)=\frac{1}{n}$ .
$\forall\frac{\epsilon}{2}>0,\exists N_{1_\epsilon}\in\mathbb{N},\forall n>N_{1_\epsilon}$, $d(x_n,0)<\frac{\epsilon}{2}$, because by Archimedean property, given epsilon, there exists a positive number $\sqrt\frac{1}{2N_{1_\epsilon}}$ where all terms higher than this number is in the open ball centered at 0 with radius epsilon.
Similarly, for $(y_n)$, for a given epsilon, you can find $\frac{1}{2N_{2_\epsilon}}$.
Denote $N^*=max\{N_1,N_2\}+2$.
Then, it follows for all epsilon greater than 0 there exists $N^*$ where for any term higher than $N^*$ the sequence $(x_n+y_n)$ converges to zero. QED.

  • $\begingroup$ 4-5. But why using that technique? ))). $\endgroup$ – kolobokish Nov 5 '16 at 23:10

Observe $\left|\dfrac{1}{n^2} - \dfrac{1}{n}\right|\le \dfrac{1}{n^2} + \dfrac{1}{n} \le \dfrac{2}{n}$. Thus choose $N \in \mathbb{N}$, such that $N \ge \dfrac{2}{\epsilon}$. So if $n > N$, then $\left|\dfrac{1}{n^2} - \dfrac{1}{n} \right| < \epsilon$, proving the limit is $0$.

  • $\begingroup$ Thanks. This was quick and shorter. $\endgroup$ – Frank Swanton Nov 5 '16 at 22:51

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