# Does a_n=lnn/(n^2+1)  converge?

Dec 13, 2015

Yes. ${\lim}_{n \to \infty} \frac{\ln \left(n\right)}{{n}^{2} + 1} = 0$

#### Explanation:

Intuitively, we should know that the given sequence converges to $0$ as any form of polynomial growth (quadratic in this case) is significantly faster than logarithmic growth. This should make sense as being the inverse of an exponential function, a logarithmic function must increase very slowly.

However, while this gives us an idea of what the result should be, it is not a proof. Let's show the result more formally. To do so, we will look at a function with the same end behavior as ${a}_{n}$.

Let $f \left(x\right) = \ln \frac{x}{{x}^{2} + 1}$.

By the construction of $f \left(x\right)$, it should be clear that if $f \left(x\right)$ converges as $x \to \infty$, then ${a}_{n}$ converges as $n \to \infty$, and ${\lim}_{n \to \infty} {a}_{n} = {\lim}_{x \to \infty} f \left(x\right)$.
(Note that this can also be proven formally with an $\epsilon - \delta$ proof, but such detail is likely unnecessary for a student of calculus).

Now, all that remains is to show that ${\lim}_{x \to \infty} f \left(x\right) = 0$.

As ${\lim}_{x \to \infty} \ln \left(x\right) = {\lim}_{x \to \infty} \left({x}^{2} + 1\right) = \infty$, we may apply L'hospital's rule to $f \left(x\right)$ as an $\frac{\infty}{\infty}$ case.

${\lim}_{x \to \infty} \ln \frac{x}{{x}^{2} + 1} = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln \left(x\right)}{\frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)}$

$= {\lim}_{x \to \infty} \frac{\frac{1}{x}}{2 x}$

$= {\lim}_{x \to \infty} \frac{1}{2 {x}^{2}}$

$= \frac{1}{\infty}$

$= 0$

And so

${\lim}_{n \to \infty} {a}_{n} = {\lim}_{x \to \infty} f \left(x\right) = 0$