# [SOLVED]Convergence 3

#### dwsmith

##### Well-known member
$\sum\limits_{n = 1}^{\infty}\left(\sqrt{1 + n^2} - n\right)$

$$\sqrt{1 + n^2} - n = \frac{1}{\sqrt{1 + n^2} + n}$$
Now what?

#### Chris L T521

##### Well-known member
Staff member
$\sum\limits_{n = 1}^{\infty}\left(\sqrt{1 + n^2} - n\right)$

$$\sqrt{1 + n^2} - n = \frac{1}{\sqrt{1 + n^2} + n}$$
Now what?
$\frac{1}{\sqrt{1+n^2}+n}\sim \frac{1}{n}$

So it makes sense to compare this to $\dfrac{1}{n}$. I'd suggest using the limit comparison test to do this and show that

$\lim_{n\to\infty}\dfrac{\dfrac{1}{n}}{\dfrac{1}{ \sqrt{1+n^2} +n}}\rightarrow L$

If the limit converges (i.e. $L<\infty$), both terms have the same behavior (in this case, the limit should converge, implying that both series diverge).

I hope this helps!

#### dwsmith

##### Well-known member
$\frac{1}{\sqrt{1+n^2}+n}\sim \frac{1}{n}$

So it makes sense to compare this to $\dfrac{1}{n}$. I'd suggest using the limit comparison test to do this and show that

$\lim_{n\to\infty}\dfrac{\dfrac{1}{n}}{\dfrac{1}{ \sqrt{1+n^2} +n}}\rightarrow L$

If the limit converges (i.e. $L<\infty$), both terms have the same behavior (in this case, the limit should converge, implying that both series diverge).

I hope this helps!
Wouldn't it be easier than to say that $\frac{1}{n} > \frac{1}{\sqrt{1+n^2}+n}$ and the Riemann Zeta function only converges for the $\text{Re}(\sigma) >1$. Since $\text{Re} (\sigma) =1$, it diverges.

#### Chris L T521

##### Well-known member
Staff member
Wouldn't it be easier than to say that $\frac{1}{n} > \frac{1}{\sqrt{1+n^2}+n}$ and the Riemann Zeta function only converges for the $\text{Re}(\sigma) >1$. Since $\text{Re} (\sigma) =1$, it diverges.
I'm not that confident with working with Zeta functions. However, I must say that in order to show divergence, you need to find a general term that is smaller than the term of the series you're looking at; that is, if you're comparing two series $\sum a_n$ and $\sum b_n$ and $a_n\leq b_n$, then $\sum b_n$ diverges if $\sum a_n$ does. So, if you want to use zeta functions, you want to find some multiple $k$ of the summand term in $\zeta(1)$ such that $\dfrac{k}{n}<\dfrac{1}{\sqrt{1+n^2}+n}$.

It's going to diverge no matter what, but you need to be careful in showing how it does so.