Tricky question Considering Divergence and Convergence

[shamieh]

Determine whether the sequence Converges or Diverges.

Tricky question, so check it out.

\(\displaystyle \frac{n^3}{n + 1}\)

So here is what I did

divided out n to get

\(\displaystyle \frac{n^2}{1} = \infty \therefore\) diverges

Now, here is what someone else did. They applied L'Hopitals, and then claimed that \(\displaystyle 3n^2 = \infty\) because \(\displaystyle 3 * \infty = \infty\) , therefore diverges.

My question is this: First of all how can you apply L'Hospitals and get that result? Isn't \(\displaystyle 3 * n^2\) still indeterminate form? Also how can you do \(\displaystyle 3 * \infty\) ? \(\displaystyle \infty\) isn't a real number, it;s like your saying \(\displaystyle 3 *\) aFakeNumber...

 

[Sudharaka]

Determine whether the sequence Converges or Diverges.

Tricky question, so check it out.

\(\displaystyle \frac{n^3}{n + 1}\)

So here is what I did

divided out n to get

\(\displaystyle \frac{n^2}{1} = \infty \therefore\) diverges

Now, here is what someone else did. They applied L'Hopitals, and then claimed that \(\displaystyle 3n^2 = \infty\) because \(\displaystyle 3 * \infty = \infty\) , therefore diverges.

My question is this: First of all how can you apply L'Hospitals and get that result? Isn't \(\displaystyle 3 * n^2\) still indeterminate form? Also how can you do \(\displaystyle 3 * \infty\) ? \(\displaystyle \infty\) isn't a real number, it;s like your saying \(\displaystyle 3 *\) aFakeNumber...

Hi shamieh, :)

Both methods are correct. What both of you are using is the Limit of the Summand Test. Note that what this tells is,

If the limit of the summand is undefined or nonzero, that is $\lim_{n\rightarrow\infty}a_n\neq 0$, then the series, $\sum_{n=0}^{\infty}a_n$ must diverge.

So in this case the limit of the summand diverges which can be taking the limit directly or by using the L'Hopital's rule (more specifically the general form of L'Hopital's rule). There is some difference between indeterminate forms and undefined forms. Indeterminate forms could be categorized as those that do not give sufficient information to evaluate the limit. For example if the limiting value of a function tends to $\frac{\infty}{\infty}$ or $\infty^0$ the L'Hopital's rule can be applied as both of these are indeterminate forms. However if the limit tends to something like $\frac{1}{0}$ or $\frac{\infty}{0}$ then L'Hopital's rule cannot be applied (both limits diverges) since these are not indeterminate forms.

I hope my clarification solves your doubts. :)

 

[Opalg]

L'Hôpital's rule deals with limits involving differentiable functions of a real variable. So strictly speaking it cannot apply to limits of sequences – you can't differentiate functions of the (discrete) set of natural numbers. If you want to apply the rule to the sequence $\dfrac{n^3}{n+1}$, you must first replace the sequence by the function $\dfrac{x^3}{x+1}$. As Sudharaka indicates, you could then apply the rule to deduce that this function diverges to $+\infty$ as $x\to\infty$, from which it follows that the sequence also diverges. But that seems like an unnecessarily roundabout way to prove this result.