I have the following series:
$\sum_{n=1}^\infty\sqrt{n^2+1} -n$
and am suppossed to proof its divergence. The ratio and root tests aren't giving me any results and it's not a telescopic series, so there must be a way to prove it by comparison.
Unfortunately though I can't think of a series that is smaller than the given series and diverges.
Could anybody help me out here?
Thanks!
On
Well,
$$\sqrt{n^2+1}-n=\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}\left(\sqrt{n^2+1}-n\right)=\frac{1}{\sqrt{n^2+1}+n}$$
and see that $\sqrt{n^2+1}+n<\sqrt{n^2+2n+1}+n=2n+1$
so that
$$\sum_{n=1}^\infty\left(\sqrt{n^2+1}-n\right)=\sum_{n=1}^\infty\frac{1}{\sqrt{n^2+1}+n}>\sum_{n=1}^\infty\frac{1}{2n+1}$$
and the last we know to diverge, thus, the initial one diverges.
Suppose you don't spot that $\sqrt{n^2+1}-n = \frac{1}{\sqrt{n^2+1}+n}$.
Then you should still estimate $\sqrt{n^2+1} - n \approx \frac1{2n}$, for example because $$\sqrt{n^2+1} = n\left(1 + \frac{1}{n^2}\right)^{1/2} = n\left(1 + \frac{1}{2n^2} + \mathcal O\left(\frac{1}{n^4}\right)\right)$$ by a Taylor series expansion or by the generalized binomial theorem.
Alternatively, because $\sqrt{n^2+1} - \sqrt{n^2}$ is approximately the derivative of $\sqrt{n^2+x}$ at $x=0$.
Once you have this estimate, we can try to prove an inequality $\sqrt{n^2+1} - n > \frac{1}{3n}$ or something like that. For this, we'd need to prove $\sqrt{n^2+1} > n + \frac{1}{3n}$, which is true (at least for all positive integers $n$) when we square both sides. Reversing the steps to turn this into a proof, we have
\begin{align} 1 &\ge \frac23 + \frac{1}{9n^2} & \text{ if }n^2 \ge \frac 13 \\ \sqrt{n^2+1} - n & \ge \sqrt{n^2 + \frac23 + \frac{1}{9n^2}} - n \\ &= \sqrt{\left(n + \frac{1}{3n}\right)^2} - n \\ &= \left(n + \frac{1}{3n}\right) - n \\ &= \frac{1}{3n} \end{align} and now a comparison with the harmonic series is enough to show divergence.