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I'm not sure if I am doing something wrong, or not... I've got an answer but it doesn't look right to me.

Given the following series, determine if it is convergent or divergent using the root or ratio test. If the test is inconclusive, use another test.

$$\sum_{n=1}^{\infty}\left(\frac{2}{e^{-8n}-1}\right)^n$$

Here's my step by step process. Maybe I left something out.

$$\lim_{n\to\infty}\left|\left(\frac{2}{e^{-8n}-1}\right)^n\right|^{\frac{1}{n}}$$ $$=\lim_{n\to\infty}\left(\frac{2}{e^{-8n}-1}\right)$$ $$=-2$$

But, even though $-2<1$, I'm very hesitant to claim that the series is convergent. I somehow feel like the answer should be positive, and I don't know if I should take the absolute value of the limit and say that the series is divergent.

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The essence of the Root Test is to compare the limit of the general term of the infinite series to the general term of a geometric series. Since we know that $ \ \Sigma_{n=1}^{\infty} a \cdot r^n \ $ converges for $ \ | r | < 1 \ $ and diverges for $ \ | r | > 1 \ $ , it is reasonable that an infinite series $ \ \Sigma_{n=1}^{\infty} (a_n)^n \ $ should converge for $ \ \lim_{n \rightarrow \infty} | a_n | < 1 \ $ and diverge where that limit of absolute value exceeds 1 . For your series, the "corresponding" geometric series is $ \ \Sigma_{n=1}^{\infty} (-2)^n \ $ , which surely diverges. –  RecklessReckoner Jun 17 '13 at 4:54

2 Answers 2

up vote 5 down vote accepted

You take the limit of the absolute value of the general term $\;\sqrt[\large n]{|a_n|}$, and after taking the $n$th root, what remains is still the absolute value of the nth root. Taking the limit of the absolute value of the nth root, in your case, will give you $$\lim_{n\to\infty}\left|\left(\frac{2}{e^{-8n}-1}\right)^n\right|^{\frac{1}{n}}\;=\;\;\lim_{n\to\infty}\left|\frac{2}{e^{-8n}-1}\right| = 2$$

Hence, in your case, the series will indeed by divergent.

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I thought the abs bars vanished after I took the n'th root of the whole thing... –  agent154 Jun 17 '13 at 4:40
    
No, not at all. It remains. The $nth$ root doesn't change the fact that what remains is absolute value of the "rooted" expression. –  amWhy Jun 17 '13 at 4:42
    
Yes, agent154: you can see the limit of the absolute value as being the absolute value of the limit. –  amWhy Jun 17 '13 at 4:43
    
I've updated my question to show my thought process... maybe you could point out where I missed something? –  agent154 Jun 17 '13 at 4:45
    
The first step after taking the nth root is mistaken: we need to keep the absolute value intact after taking the nth root. That's crucial when you test for absolute convergence, for example, of an alternating series. –  amWhy Jun 17 '13 at 4:48

You absolutely need toconsider $|u_n|$ under the $$\sqrt[n]{...}$$ because the test tells us to do $$\lim_{x\to\infty}\sqrt[n]{|u_n|}$$ And that's why you would find the series divergent.

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Here's another question - is $\lim_{n\to\infty}|a_n|=|\lim_{n\to\infty} a_n|$? Otherwise I have no idea how things change. –  agent154 Jun 17 '13 at 4:39
    
@agent154: The function $|...|$ is a continuous function. –  Babak S. Jun 17 '13 at 4:42
2  
Be careful. Certainly it can happen that $\lim |a_n|$ exists and $\lim a_n$ does not! –  Ted Shifrin Jun 17 '13 at 5:03
    
@TedShifrin: Yes Prof. I just noted this matter for this series. Of course $\sum\frac{(-1)^n}{n}$ and $\sum\frac{1}{n}$ have a notable behaviour. Thanks –  Babak S. Jun 17 '13 at 5:25

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