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Evaluate $$\lim_{n\to \infty}\dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+2n}}$$

$$$$ I came across the the question on this site itself but had a few doubts on the given solution. As I do not yet have 50 reputation points, I cannot comment over there. Could somebody please help me? $$$$ From what I understand of the Squeeze Theorem, the three functions are related as $$g(x)\le f(x)\le h(x)$$ $$$$ Now in the selection of terms, the following inequality has been used: $$\displaystyle \frac{1}{n+1} \leq \frac{1}{\sqrt{n^2+k}} \leq \frac{1}{n} $$when $0 \leq k \leq 2n$ $$$$ This inequality lead to the one used as the three functions for the application of the Squeeze theorem:

$$\frac{2n+1}{n+1} \leq S(n) \leq \frac{2n+1}{n}$$ where $S(n)=\dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+2n}}$ $$$$ I don't understand how $$\displaystyle \frac{1}{n+1} \leq \frac{1}{\sqrt{n^2+k}} $$ $$$$ Isn't $$n^2+k\le n^2+2n<(n+1)^2 \Rightarrow n^2+k<(n+1)^2$$ $$\Rightarrow \sqrt{n^2+k}< (n+1)$$$$\Rightarrow \displaystyle \frac{1}{n+1} < \frac{1}{\sqrt{n^2+k}}$$

$$$$ Thus shouldn't the resulting set of inequalities be $$\displaystyle \frac{1}{n+1} < \frac{1}{\sqrt{n^2+k}} \leq \frac{1}{n} $$

$$\Longrightarrow \frac{2n+1}{n+1} < S(n) \leq \frac{2n+1}{n}$$

$$$$

In this case there is a $<$ sign instead of the $\le$ sign. How then can the Squeeze Theorem be applied? Many thanks in advance. $$$$ EDIT: Also, since Limits preserve Inequalities, how can $$\lim_{n\to \infty} \frac{2n+1}{n+1} = \lim_{n\to \infty}\frac{2n+1}{n}$$ when $$\frac{2n+1}{n+1} < \frac{2n+1}{n}$$

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    $\begingroup$ Well $0<1/n$ for all $n.$ Yet $\lim 1/n = 0.$ $\endgroup$ – zhw. May 31 '16 at 1:35
  • $\begingroup$ Also note $a<b \implies a\le b.$ $\endgroup$ – zhw. May 31 '16 at 1:38
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    $\begingroup$ The last term is $\frac{1}{\sqrt{n^2+2n}}$, which is greater than $\frac{1}{\sqrt{n^2+2n+1}}=\frac{1}{n+1}$, and all the terms before that are even bigger. $\endgroup$ – André Nicolas May 31 '16 at 1:40
  • $\begingroup$ @zhw. True, but then I had read that Limits always preserve Inequalities in several contexts, whereas this doesn't seem to be the case. Is there a specific framework within which the Inequalities are preserved? $\endgroup$ – user342209 May 31 '16 at 1:40
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    $\begingroup$ They do not preserve strict inequality. $\endgroup$ – André Nicolas May 31 '16 at 1:41
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Limits don't necessarily preserve strict inequalities. For example, $1-\frac{1}{n}<1+\frac{1}{n}$, yet they have the same limit as $n$ goes to $\infty$.

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Remember that $a<b\implies a\le b$. This is because $a\le b$ means $a<b$ or $a=b$ as you already noted in the comment.

If you are still confused, recall that "$p\implies q$" means $q$ is true whenever $p$ is true. And when $a<b$ is true, $a<b$ or $a=b$ is true. Hence, $a<b\implies a\le b$.

For example, is "$0\le 1$" true?

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  • $\begingroup$ But doesn't $a<b$ (ie a is less than b) rule out $a\le b$ (ie a is less than or equal to b)?$$$$ I thought so since $a<b$ necessarily means that $a$ is $smaller$ than $b$, and thus $cannot$ be $equal$ to b (as it is smaller than b). $\endgroup$ – user342209 Jun 1 '16 at 6:54
  • $\begingroup$ Of course $a<b$ rules out $a=b$ but not $a\le b$. $\endgroup$ – Eclipse Sun Jun 1 '16 at 15:47

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