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- Thread starter dwsmith
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- Jan 26, 2012

- 995

\[\frac{1}{\sqrt{1+n^2}+n}\sim \frac{1}{n}\]$\sum\limits_{n = 1}^{\infty}\left(\sqrt{1 + n^2} - n\right)$

$$

\sqrt{1 + n^2} - n = \frac{1}{\sqrt{1 + n^2} + n}

$$

Now what?

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!

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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.\[\frac{1}{\sqrt{1+n^2}+n}\sim \frac{1}{n}\]

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!

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- #4

- Jan 26, 2012

- 995

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 isWouldn'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.

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