Evaluate $$\lim_{n\to \infty}\dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+2n}}$$
$$$$ I'm supposed to solve this problem using the Squeeze Theorem. I had selected the functions:$$$$$$\dfrac{1}{\sqrt{n^2}}\le \dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+2n}}\le \dfrac{1}{\sqrt{n^2}}+...+\dfrac{1}{\sqrt{n^2+2n}}+\dfrac{1}{\sqrt{n^2+2n+1}}$$
$$$$I know that these functions that I had selected are wrong because they are strictly less than or strictly more than the given function in the limit. I can't understand what functions to choose which would maintain the $\le$ sign. $$$$ Could somebody please show me what functions to select instead? Many thanks!
Your sum $S(n)$ has
$$\frac{2n+1}{n+1} \leq S(n) \leq \frac{2n+1}{n}$$
Can you see why?
Hint: $\displaystyle \frac{1}{n+1} \leq \frac{1}{\sqrt{n^2+k}} \leq \frac{1}{n}$ when $0 \leq k \leq 2n$.
Hint 2: $n^2+k \leq n^2+2n$ (for $k \leq 2n$)
Hence $ n^2+k < n^2+2n+1 = (n+1)^2$. Hence $\sqrt{n^2+k} \leq n+1$.
Hint 3: There are $2n+1$ terms.
Taking the limit gives
$$\lim_{n \to \infty}S(n)=2$$