Understanding squeeze theorem

Question

$$ \lim_{x \to +\infty} \frac{1}{\sqrt{(x^2)}} + \frac{1}{\sqrt{(x^2+1)}} + \frac{1}{\sqrt{(x^2+2)}}......+ \frac{1}{\sqrt{(x^2+2x)}}$$ This is the given problem now someone showed me that this could be solved using squeeze theorum $$\frac{2x+1}{\sqrt{x^2+2x}}\leqslant S_x\leqslant\frac{2x+1}{\sqrt{x^2}},$$ which is enough to deduce that $\lim\limits_{x\to+\infty}S_x=2$. now i understand what the theorem is but i have problem in understanding how it was applied. How do we know the range of this series , how the inequality terms were chosen?

Answer 1

The series is $\sum_{k=0}^{2x} \frac{1}{\sqrt{x^2+k}}$, which assumes $x$ is a positive integer and so there are $2x+1$ terms in this series, all of which are decreasing as $k$ gets larger. So the lower bound is just the smallest term: $\frac{1}{x^2+2x}$ times the number of terms which is $2x+1$. Hopefully you can now work out how to get the upper bound.