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Find $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}$.

I claim that $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}=2$. To prove this, for given $\varepsilon >0$, I have to find $M\in N$ such that $|\frac {2n^2+10n+5}{n^2}-2|<\varepsilon$ for $n \ge M$.

By Archimedean property, we can find $M \in N$ such that $\frac {15}M<\varepsilon$, and note that $n\ge M \rightarrow \frac 1n \le \frac 1M \rightarrow \frac {15}n \le \frac {15}M$.

Then, for $n \ge M$, we have that $|\frac {2n^2+10n+5}{n^2}-2|=|\frac {10n+5}{n^2}| < |\frac {15n}{n^2}|$ (since $n \ge M \in N$) $<\frac {15}n\le \frac {15}M<\varepsilon$.

Therefore, by definition of convergence, $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}=2$.

Should I say $M\in Z^+$ (because I am worried about the case where $M=0$)? Can you find any mistakes in this proof?

Thank you in advance.

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    $\begingroup$ Your proof is fine. $\endgroup$ – DeepSea Apr 15 '18 at 7:19
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Your proof is correct, well done.

Some authors don't include $0$ in $\mathbb{N}$, but based on your question seems $\mathbb{N}=\{0,1,2,\dots\}$ for you.

I guess the definition of convergence of a sequence $(u_n)_{n\in\mathbb{N}}$ to $l\in\mathbb{R}$ that you have is this:

$$\forall\varepsilon>0,\exists M\in\mathbb{N},\forall n\ge M,|u_n-l|<\varepsilon.$$

Let $\varepsilon>0$ and $M\in\mathbb{N}$ (that exists) such that $$\forall n\ge M,|u_n-l|<\varepsilon.$$

Notice that, in particular, $$\forall n\ge M+1,|u_n-l|<\varepsilon$$ because $M+1>M$. So you can replace $M$ by $M+1$ or any larger integer. Therefore you can always choose $M$ to be positive. So you can safely have this as a definition:

$$\forall\varepsilon>0,\exists M\in\mathbb{Z}^+,\forall n\ge M,|u_n-l|<\varepsilon.$$

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Your proof is fine, to simplify note that

$$\frac {2n^2+10n+5}{n^2}=2+\frac{10}{n}+\frac5{n^2}$$

then it suffices to prove that $\frac1n \to 0$.

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  • $\begingroup$ Dear gimusi, can you check my solution? If you have a few minutes.. Is it possible..? Thank you soo much. math.stackexchange.com/q/3043357/460967 $\endgroup$ – Elvin Dec 17 '18 at 21:43
  • $\begingroup$ @Student That's not the topic where I'm most confident but it seems a nice solution to me! Maybe you should write it in mathjax to encourage more answers! Or as an alternative you could give a small bounty to give more attention to that! $\endgroup$ – user Dec 17 '18 at 21:51
  • $\begingroup$ What do you use to write in Latex? I use Android keyboard. This is so difficult..:( $\endgroup$ – Elvin Dec 18 '18 at 9:23
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Another possibility: apply two times Cesàro-Stolz. $$ \lim_{n\to\infty}\frac{2n^2 + 10n + 5}{n^2} = \lim_{n\to\infty}\frac{((2(n+1)^2 + 10(n+1) + 5) - (2n^2 + 10n + 5)}{(n+1)^2 - n^2} = \lim_{n\to\infty}\frac{4n + 12}{2n + 1} = \lim_{n\to\infty}\frac{(4(n+1) + 12) -(4n + 12)}{(2(n+1) + 1) - (2n + 1)} = \lim_{n\to\infty}\frac42 = 2. $$

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