Choosing Functions for the Squeeze Theorem 
Evaluate $$\lim_{n\to \infty}\dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+2n}}$$

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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?
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From what I understand of the Squeeze Theorem, the three functions are related as 
$$g(x)\le f(x)\le h(x)$$
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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 $0 \leq k \leq 2n$
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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 $S(n)=\dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+2n}}$
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I don't understand how $$\displaystyle \frac{1}{n+1} \leq \frac{1}{\sqrt{n^2+k}} $$
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Isn't $$n^2+k\le n^2+2n<(n+1)^2 \Rightarrow n^2+k<(n+1)^2$$
$$\Rightarrow \sqrt{n^2+k}< (n+1)$$$$\Rightarrow \displaystyle \frac{1}{n+1} < \frac{1}{\sqrt{n^2+k}}$$
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Thus shouldn't the resulting set of inequalities be $$\displaystyle \frac{1}{n+1} < \frac{1}{\sqrt{n^2+k}} \leq \frac{1}{n}  $$
$$\Longrightarrow \frac{2n+1}{n+1} < S(n) \leq \frac{2n+1}{n}$$
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In this case there is a $<$ sign instead of the $\le$ sign. How then can the Squeeze Theorem be applied?
Many thanks in advance.
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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 $$\frac{2n+1}{n+1} <  \frac{2n+1}{n}$$

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 $\infty$.  
