L'Hopital rule for sequence?

Question

I had a task to calculate divergence of $$ \sum_{n=0}^{\infty} \frac{3^n}{2n^2+5}x^n $$ During that task I had to calculate the following limit: $$ lim_{n\rightarrow \infty}\ \frac{ln(2n^2+5)}{n} $$ so naturally I used the L'Hopital rule, but my teacher said that I can't differentiate sequence.

My question: how can I calculate it then? Would making a function $f(x)=\frac{ln(2x^2+5)}{x}$ and calculating it's limit with L'Hopital and then saying that it's limit is the same as the sequence's limit would be a correct solution?

Answer 1

Here is the proof that doesn't use L'Hospital, but it uses sandwich theorem.

$$\lim_{n \to \infty}{\frac{\ln{(2n^3+5)}}{n}} $$

First we will substitute $z=2n^3+5$

$$\lim_{z \to \infty}{\frac{\ln{z}}{\sqrt[3]{\frac{z-5}{2}}}} $$

Now divide by $\sqrt[3]{z}$ in the numerator and denominator.

We get;

$$\lim_{z \to \infty}{\frac{\frac{\ln{z}}{\sqrt[3]{z}}}{\sqrt[3]{\frac{z-5}{2z}}}} $$

Now we can evaluate the limit of numerator and denominator independently since they both exist.

First the denominator:

$$\lim_{z \to \infty}{{\sqrt[3]{\frac{z-5}{2z}}}}=\lim_{z \to \infty}{{\sqrt[3]{\frac{1}{2}-\frac{5}{2z}}}}=\sqrt[3]{\frac{1}{2}}$$

And the numerator:

$$\lim_{z \to \infty}{\frac{\ln{z}}{\sqrt[3]{z}}}$$

For this we will use sandwich theorem. Note that

$$\ln{z}<z$$ as $z$ approaches infinity so:

$$\frac{\ln{z}}{\sqrt[3]{z}}=\frac{\ln{\sqrt[4]{z}^4}}{\sqrt[3]{z}}=\frac{4\ln{\sqrt[4]{z}}}{\sqrt[3]{z}}<\frac{4\sqrt[4]{z}}{\sqrt[3]{z}}$$

While we can also say that:

$$\frac{\ln{z}}{\sqrt[3]{z}}>0$$ since it is positive as $z$ approaches infinity.

We showed that: $$0<\frac{\ln{z}}{\sqrt[3]{z}}<\frac{4\sqrt[4]{z}}{\sqrt[3]{z}}$$

We can easily evaluate these two limits: $$\lim_{z \to \infty}{0}=0 \ and \lim_{z \to \infty}{\frac{4\sqrt[4]{z}}{\sqrt[3]{z}}}=0$$

So, by the sandwich theorem we can conclude that: $$\lim_{z \to \infty}{\frac{\ln{z}}{\sqrt[3]{z}}}=0$$

Our overall limit is now:

$$\lim_{z \to \infty}{\frac{\frac{\ln{z}}{\sqrt[3]{z}}}{\sqrt[3]{\frac{z-5}{2z}}}}=\frac{0}{\sqrt[3]{\frac{1}{2}}}=0 $$

Answer 2

You can perfectly apply L'Hopital's Rule to compute the limit $\lim_{n\to\infty}\frac{\log(2n^2+5)}n$; you will get that it is equal to $0$.

Having said this, I have no idea why you would want to do that to study that power series. All you have to do is to compute\begin{align}\lim_{n\to\infty}\frac{\left\lvert\frac{3^{n+1}}{2(n+1)^2+5}x^{n+1}\right\rvert}{\left\lvert\frac{3^n}{2n^2+5}x^2\right\rvert}&=\lim_{n\to\infty}\frac{2n^2+5}{2(n+1)^2+5}3\lvert x\rvert\\&=3\lvert x\rvert.\end{align}So, your series diverges when $\lvert x\rvert<\frac13$ and diverges when $\lvert x\rvert>\frac13$. Actually, it follows from the comparison test that it converges when $\lvert x\rvert\leqslant\frac13$.