# Does the sequence converge?

#### evinda

##### Well-known member
MHB Site Helper
Hey!!!
I want to check if the sequence $a_{n}=\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}$ converges.
I thought that I could find the difference $a_{n+1}-a_{n}$ to check if $a_{n}$ is increasing or decreasing.I found:
$a_{n+1}-a_{n}=\sum_{i=1}^{n}(\frac{1}{\sqrt{(n+1)^{2}+i)}}-\frac{1}{\sqrt{n^2+i}})+\frac{1}{\sqrt{(n+1)^{2}+n+1}}$..But from that we cannot conclude if the difference is negative or positive,right?? So,what else could I do??

#### Plato

##### Well-known member
MHB Math Helper
Hey!!!
I want to check if the sequence $a_{n}=\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}$ converges.
Is it the case that $$\frac{n}{{\sqrt {{n^2} + n} }} \le {a_n} \le 1$$?

Is $${a_n}$$ an increasing sequence?

#### evinda

##### Well-known member
MHB Site Helper
Is it the case that $$\frac{n}{{\sqrt {{n^2} + n} }} \le {a_n} \le 1$$?

Is $${a_n}$$ an increasing sequence?
I found this:$\frac{n}{\sqrt{n^2+n}}\leq a_{n}\leq \frac{n}{\sqrt{n^2+1}}$..
So,could I just say that from the squeeze theorem the limit is $1$,without finding the monotonicity??

#### evinda

##### Well-known member
MHB Site Helper
I found this:$\frac{n}{\sqrt{n^2+n}}\leq a_{n}\leq \frac{n}{\sqrt{n^2+1}}$..
So,could I just say that from the squeeze theorem the limit is $1$,without finding the monotonicity??
Or can't I do it like that,because it is not given that the sequence converges??

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I found this:$\frac{n}{\sqrt{n^2+n}}\leq a_{n}\leq \frac{n}{\sqrt{n^2+1}}$..
So,could I just say that from the squeeze theorem the limit is $1$,without finding the monotonicity??
Yep. That works. Monotonicity not required.

#### evinda

##### Well-known member
MHB Site Helper
Yep. That works. Monotonicity not required.
Great!!!!Thank you very much!!!!