calculate the limit of a numerical sequence

Question

$(Sn)= \sum_{i=1}^{2n+1} \frac{1}{\sqrt{n^{2}+i}}$

I need to proove that the limit of this sequence equals $2$ but i'm stuck here. i tried give the sequence upper and lower bounds to use " gendarme" but it doesn't seem to work. any help would be appreciated

Answer 1

Each term is between $1/\sqrt{n^2+1}$ and $1/\sqrt{n^2+2n+1}$ or $1/n$ and $1/(n+1)$.

Answer 2

Since each term $\frac{1}{\sqrt{n^2+i}} \geq \frac{1}{\sqrt{n^2+2n+1}} = \frac{1}{n+1}$ and we have $2n$ terms, we have a lower bound of $\lim_{n\to\infty}\frac{2n}{n+1} = 2$.

We also have for each term $\frac{1}{\sqrt{n^2+i}} \leq \frac{1}{n}$. Again, we have $2n$ terms and so we get an upper bound of $\lim_{n\to\infty}\frac{2n}{n} = 2$.

Thus by the Squeeze Theorem (or "gendarme") the limit must equal 2.

Answer 3

More generally, if $S_n =\sum_{i=1}^{an+b} \dfrac1{\sqrt{n^2+i}} $, then

$\begin{array}\\ S_n &=\sum_{i=1}^{an+b} \dfrac1{\sqrt{n^2+i}}\\ &\lt\sum_{i=1}^{an+b} \dfrac1{\sqrt{n^2}}\\ &=\sum_{i=1}^{an+b} \dfrac1{n}\\ &=\dfrac{an+b}{n}\\ &=a+\dfrac{b}{n}\\ &\to a\\ \text{and}\\ S_n &=\sum_{i=1}^{an+b} \dfrac1{\sqrt{n^2+i}}\\ &\gt\sum_{i=1}^{an+b} \dfrac1{\sqrt{n^2+an+b}}\\ &\gt\sum_{i=1}^{an+b} \dfrac1{n\sqrt{1+a/n+b/n^2}}\\ &=\sum_{i=1}^{an+b} \dfrac1{n(1+a/(2n)+b/(2n^2))} \qquad\text{since } \sqrt{1+x} < 1+x/2\\ &=\dfrac{an+b}{n+a/2+b/(2n)}\\ &=\dfrac{a+b/n}{1+a/2n)+b/(2n^2)}\\ &\gt(a+b/n)(1-a/(2n)-b/(2n^2)) \qquad\text{since }1/(1+x) > 1-x\\ &=a+(b-a/2)/n-ab/n^2-b^2/(2n^3)\\ &\to a\\ \end{array} $

so $S_n \to a$.