Show that the sequence $((\frac n {1+n^2} + \frac 1 3)^n)_{n\in\Bbb N}$ is a null sequence.

Question

I've been asked to show the sequence is a null sequence.

$$\left( \left( \frac n {1+n^2} + \frac 1 3\right)^n\right)_{n\in\Bbb N}$$

I previously have used L'Hopital's rule when attempting to solve such problems, but this does not fit the conditions necessary to apply it.

How should one go about tackling any sequence convergence problem when the sequence is raised to the exponent n?

Answer 1

The argument for showing that the sequence converges to $0$ is much easier here.

Since for some $a$ with $|a|<1$ it holds $a^n \rightarrow 0$, it suffices to show that $0<\frac{n}{1+n^2}+\frac{1}{3} < 1$. The first inequality is clear and the second is equivalent to $\frac{n}{1+n^2} < \frac{2}{3}$. And since $\frac{n}{1+n^2}$ is clearly a null sequence, for $n$ large enough the latter inequality holds.

Answer 2

Show that $$0<\left(\frac{n^2+3n+1}{3(n^2+1)}\right)^n<1$$

Answer 3

For $n \geq 3$, it's trivial to show that $\dfrac{n}{1+n^2} < \dfrac{1}{3}$.

Thus we know $\dfrac{n}{1+n^2} + \dfrac{1}{3} < \dfrac{2}{3}$. Can you finish from here?

Answer 4

Notice that, for $n \gt 1$ $$ \frac{n}{1+n^2} \lt \frac{1}{2} $$ and then $$ \frac{n}{1+n^2} + \frac{1}{3} \lt \frac{5}{6} $$

what's the limit of $\left(\frac{5}{6}\right)^n$ ?