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Raabe's test, logarithm test and Bertrand test are the most commonly used criterion in calculus. The relationship between them is quite interesting. Here is how:

$\sum\limits_{n=1}^\infty a_n$, $a_n\gt0$,

1). If $\lim\limits_{n\to\infty}n(\dfrac{a_n}{a_{n+1}}-1)=\alpha$, then

$$\lim_{n\to\infty} \frac{\ln\frac1{a_n}}{\ln n}=\alpha $$

2). If $\lim\limits_{n\to\infty}\ln n\left[n\left(\dfrac{a_n}{a_{n+1}}-1\right)-1\right]=\beta$, then

$$\lim_{n\to\infty} \frac{\ln\frac1{na_n}}{\ln\ln n}=\beta $$

I have no clue about it. I don't know how to prove it. Could anyone help me? Thanks a lot

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The only question here is wether someone can help you. Help you with what? –  Git Gud Jan 4 at 16:26
    
@amWhy show no effort? I can do nothing for some problems –  ziang chen Jan 5 at 16:46
    
for example, math.stackexchange.com/questions/624251/… –  ziang chen Jan 5 at 16:48
    
@amWhy My mother tongue is not English, so I can't write many sentences –  ziang chen Jan 5 at 16:56
    
ziang: I deleted my comment. When you are unable to even begin a problem, perhaps you should say so, but ask for help getting started, or for a hint. Otherwise, it just looks like you're just seeking answers, rather than seeking help to get started. By the way, if it's any consolation, I think you've been doing a fine job with your English! ;-) –  amWhy Jan 5 at 17:03
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1 Answer

I guess Stolz Cesaro coupled with $\lim_{x \to 0}(1+x)^{1/x}=e$ should do the trick.

We have that $\lim_{n\to \infty}n (\frac{a_n}{a_{n+1}}-1) =\alpha$ and we want to calculate $\lim_{n \to \infty} \frac{-\ln a_n}{\ln n}$. Since we want to apply Stolz Cesaro, we look at the limit of $$ L= \lim_{n \to \infty} \frac{\ln a_n -\ln a_{n+1}}{\ln(n+1)-\ln n}=\lim_{n \to \infty} \frac{\ln\frac{ a_n}{ a_{n+1}}}{\ln \frac{n+1}{n}} \ (1)$$

Note that if $\alpha$ is finite (which will be my assumption) then $a_n/a_{n+1}$ goes to $1$ so we have $$ \lim_{n \to \infty} (\frac{a_n}{a_{n+1}})^{n}=\lim_{n \to \infty}\left((1+\frac{a_n}{a_{n+1}}-1)^{\frac{1}{(a_n/a_{n+1}-1)}}\right)^{n(\frac{a_n}{a_{n+1}}-1)}=e^\alpha$$

To calculate $L$, multiply both numerator and denominator in $(1)$ with $n$ and use the above limit and $\lim_{n\to \infty} (1+\frac{1}{n})^n=e$.

The second limit could be tackled in a similar way (but it clearly gets more complicated...). This is a nice result, and it proves at once the Raabe test, that if $\alpha>1$ then the series $\sum a_n$ is comparable with the harmonic series $\sum \frac{1}{n^\alpha}$ and therefore is convergent.

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