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Does the sequence converge?

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
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
Jan 27, 2012
196
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 [tex]\frac{n}{{\sqrt {{n^2} + n} }} \le {a_n} \le 1[/tex]?

Is [tex]{a_n}[/tex] an increasing sequence?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Is it the case that [tex]\frac{n}{{\sqrt {{n^2} + n} }} \le {a_n} \le 1[/tex]?

Is [tex]{a_n}[/tex] 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?? :confused:
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
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?? :confused:
Or can't I do it like that,because it is not given that the sequence converges?? :confused:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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?? :confused:
Yep. That works. Monotonicity not required.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720