0
$\begingroup$

Based on the theorem below, when using the integral test to prove the convergence or divergence of a series, does one need to also prove the series itself is decreasing, continuous and positive? Would you use Induction for each?

I'm currently trying to prove that $$\sum\limits_{n=2}^\infty \frac{1}{n(\ln(n))^p}$$ converges if $p\gt 1$ and diverges if $0\lt p \lt 1$. Which I think I have done. My proof is something similar to the answer here.

So, I want to fully prove the series converges, but at the same time, I don't really want to prove a bunch of aspects about the function if they aren't strictly required as part of the proof.

Any tips would be appreciated.


The integral test theorem

$\endgroup$
1
$\begingroup$

The requirements are necessary for the theorem to work.

To prove that they are fulfilled in your case, induction won't be the way to go. In fact, it is much easier. You define

$$f(x) = \frac{1}{x(\ln{x})^p}$$

for $x\geq 2$.

Is $f$ continuous? Yes, it is composed of continuous functions.

Is $f$ positive? Yes, we have that $x(\ln{x})^p>0$ for $x\geq 2$.

And is $f$ decreasing? It is, because both $x$ and $(\ln{x})^p$ are increasing functions, thus so is $x(\ln{x})^p$, and consequently $f(x)$ is decreasing.

$\endgroup$
1
  • $\begingroup$ Thanks. This really helps. $\endgroup$
    – Akyidrian
    Mar 31 '15 at 11:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.