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Does the sequence diverge?

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hey again!!! (Blush)
I have to check if the sequence $a_{n}=\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+...+\frac{n}{\sqrt{n^2+n}}$ converges.I thought that:$$\frac{n^{2}(n+1)}{2\sqrt{n^2+n}} \leq a_{n} \leq \frac{n^{2}(n+1)}{2\sqrt{n^2+1}}$$ Because of the fact that:
$$\lim_{n \to \infty}\frac{n^{2}(n+1)}{2\sqrt{n^2+n}}=\lim_{n \to \infty}\frac{n^{2}(n+1)}{2\sqrt{n^2+1}}=\infty$$ I though that the sequence diverges.
Could you tell me if it is right?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Hey again!!! (Blush)
I have to check if the sequence $a_{n}=\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+...+\frac{n}{\sqrt{n^2+n}}$ converges.I thought that:$$\frac{n^{2}(n+1)}{2\sqrt{n^2+n}} \leq a_{n} \leq \frac{n^{2}(n+1)}{2\sqrt{n^2+1}}$$ Because of the fact that:
$$\lim_{n \to \infty}\frac{n^{2}(n+1)}{2\sqrt{n^2+n}}=\lim_{n \to \infty}\frac{n^{2}(n+1)}{2\sqrt{n^2+1}}=\infty$$ I though that the sequence diverges.
Could you tell me if it is right?
Right idea, but each time that you have written $n^2$ it ought to be just $n$ (not squared).
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Right idea, but each time that you have written $n^2$ it ought to be just $n$ (not squared).
Why?? Istn't $1+2+....+n=\frac{n(n+1)}{2}$ and we take for the left inequality $n$ times $\frac1{\sqrt{n^2+n}}$ and for the second $n$ times $\frac{1}{\sqrt{n^2+1}}$ ??Or am I wrong?? (Thinking)
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Why?? Isn't $1+2+....+n=\frac{n(n+1)}{2}$ and we take for the left inequality $n$ times $\frac1{\sqrt{n^2+n}}$ and for the second $n$ times $\frac{1}{\sqrt{n^2+1}}$ ?? Or am I wrong?? (Thinking)
Take it more slowly: $$\begin{aligned}a_{n} &=\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+ \ldots +\frac{n}{\sqrt{n^2+n}} \\ &\geqslant \frac{1}{\sqrt{n^2+n}}+\frac{2}{\sqrt{n^2+n}}+ \ldots +\frac{n}{\sqrt{n^2+n}} \\ &= \frac{1}{\sqrt{n^2+n}}(1+2+\ldots +n) = \frac{n(n+1)}{2\sqrt{n^2+n}}. \end{aligned}$$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Take it more slowly: $$\begin{aligned}a_{n} &=\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+ \ldots +\frac{n}{\sqrt{n^2+n}} \\ &\geqslant \frac{1}{\sqrt{n^2+n}}+\frac{2}{\sqrt{n^2+n}}+ \ldots +\frac{n}{\sqrt{n^2+n}} \\ &= \frac{1}{\sqrt{n^2+n}}(1+2+\ldots +n) = \frac{n(n+1)}{2\sqrt{n^2+n}}. \end{aligned}$$
Ok..I understand...So,can I use the relation:
$$\frac{n(n+1)}{2\sqrt{n^2+n}} \leq a_{n} \leq \frac{n(n+1)}{2\sqrt{n^2+1}}$$ to conclude that the sequence diverges??