I am suppose to use the limit comparison test to prove divergence or convergence. There isn't really any examples in my book that show how to pick your $b_n$ so I just pick whatever works out nicely.

$$a_n = \dfrac{1}{\sqrt{n} + \ln n}$$

$$b_n = \dfrac{1}{n^4}$$

I chose this because I know that it converges and that it makes the math easier.

$$\dfrac {\dfrac{1}{\sqrt{n} + \ln n}}{\dfrac{1}{n^4}}$$

$$ {\dfrac{n^4}{\sqrt{n} + \ln n}}$$

So now I look at my Table of Truths to find the answer.

I know that my $b_n$ converges, that is easy. It goes to zero.

My $a_n$ I am not sure about. I do know that the limit of $\dfrac{a_n}{b_n}$ is greater than zero.

According to the table it converges if $b_n$ does, but infinity is another option so I guess that this is really infinity which is only converges if $a_n$ does. So now I need to look at $a_n$ but really isn't an infinite limit > 0? Besides that logical confusion $a_n$ does converge to zero so that means $b_n$ converges as well. Which I guess is kind of confusing to me, because it seems like I didn't really prove anything since I just looked at the original function which I know goes towards zero anyways.

What just happened? I feel like I did nothing.



If you're fairly new to sequences and series, a quick and "dirty" way to show that the series diverges (it fails to converge) is to compare it to $b_n = \dfrac 1n.\;$ Recall, the harmonic series diverges: $\sum_{n\to \infty} \frac 1n \to \infty$.

And, you can confirm: $$\forall n \geq 3,\quad \dfrac 1n \leq \dfrac{1}{\sqrt n + \ln n}$$

  • $\begingroup$ So I am suppose to know that logn is less than n and lnn * square root n is still less than n? $\endgroup$ – Paul the Pirate Jul 18 '13 at 0:58
  • $\begingroup$ Yes, it sure is: the product is equal to n only for $n = 1$. All you need to know is that $(\ln n) \leq \sqrt n$, since $\sqrt n\cdot \sqrt n = n$, so $\sqrt n\cdot \ln n \leq n$ $\endgroup$ – Namaste Jul 18 '13 at 1:01
  • $\begingroup$ Well what if I wanted to actually evaluate this sequence? What is a good $b_n$ to pick and why? Can it by any positive number? $\endgroup$ – Paul the Pirate Jul 18 '13 at 1:06
  • $\begingroup$ See this graph, Paul: square root vs ln $\endgroup$ – Namaste Jul 18 '13 at 1:07
  • $\begingroup$ Tackle convergence, divergence first. If a series diverges, it diverges to infinity (there's no finite sum to be had!). $b_n = \dfrac 1n$ is a great choice. $\endgroup$ – Namaste Jul 18 '13 at 1:10

Do you have not noticed that $$a_n\sim_\infty \frac{1}{\sqrt n}?$$


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