1
$\begingroup$

The series $$\sum_{n=1}^{\infty} \frac{1-e}{e^{n}}$$ when subjected to the geometric series convergence test, is convergent:

$$\sum_{n=1}^{\infty} \frac{1-e}{e^{n}} = \sum_{n=1}^{\infty} (1-e)(\frac{1}{e})^{n}$$ and since $\frac{1}{e} < 1$ it is convergent, but when subjected to D'Alembert's test it is divergent:

$$\lim_{n\to\infty}\mid\frac{\frac{1-e}{e^n}}{\frac{1-e}{e^{n+1}}}\mid = e$$ and since $e > 1$ it's divergent.

I'm sure I'm messing up somewhere but I just can't seem to find it and it's driving me nuts.

Edit: The series is clearly convergent, I'm only trying to find out what I'm doing wrong (definitely in my application of D'Alembert's criterion).

Edit2: Oh ffs! I'm tired as hell but I don't know how I just didn't spot that. Thanks, I'd have accepted an answer already if I could!

$\endgroup$
  • 1
    $\begingroup$ You are wronfully write D'Alambert criterion for series: $\lim \frac{a_{n+1}}{a_n}$, not $a_n/a_{n+1}$. $\endgroup$ – Michael Galuza Jun 5 '15 at 17:52
  • $\begingroup$ Note that the ratio test is essentially a limit comparison test against a geometric series (and the test itself tells you what series to compare to). Thinking of it that way should help you avoid making a mistake like this. $\endgroup$ – Ian Jun 5 '15 at 17:58
0
$\begingroup$

You've got the d'Alembert test the wrong way around -- you're computing $\left|\frac{a_n}{a_{n+1}}\right|$ instead of the correct $\left|\frac{a_{n+1}}{a_{n}}\right|$.

$\endgroup$
  • $\begingroup$ I knew it was something really really basic but I just wasn't seeing it. Thanks! $\endgroup$ – HolyThunder Jun 5 '15 at 17:55
3
$\begingroup$

You're doing the ratio test upside down. We want to look at

$$L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$$

You put $a_{n+1}$ in the denominator. Invert your result and you get $L = \frac1e < 1$, so the series converges absolutely.

$\endgroup$
  • $\begingroup$ I'm really tired, for some reason I absolutely did not notice that. Thanks! $\endgroup$ – HolyThunder Jun 5 '15 at 17:55
0
$\begingroup$

I'd like to add something. The entire reason why the ratio test exists, the entire reason why we even consider the ratio test is because of geometric series! What the ratio test is telling you is that for big $n$, $a_n$ is APPROXIMATELY geometric, so it will behave the same way. What we have here is something much, much stronger! We don't have that $a_n$ tends to something "approximately" geometric. We have that $a_n$ is EXACTLY geometric. So the ratio test HAS to give you $\frac1e$!

$a_n$ being geometric is MUCH more fundamental than the ratio test.

$\endgroup$
  • $\begingroup$ Yeah, I actually came to that conclusion while writting this up. Although in the post I wrote the geometric series test first, I actually tried the ratio test first. I knew that that result has to be the same, and that the results being inverse weren't a coincidence but I jsut wan't seeing what I was doing wrong. However, I totally understand what you mean. The ratio test is like a broader geometric series test. $\endgroup$ – HolyThunder Jun 5 '15 at 18:13

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.