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I can't find a way to test the convergence/divergence of this series:

$$ \sum_{n=2}^{\infty}\frac{n^4}{\log(1)+\log(2)+\log(3)+\cdots+\log(n)} $$

I tried the Cauchy method but in order to make the logarithms more manageable I grouped them all (so $\log(1)+\log(2)+\log(3)+...+\log(n)=\log(n!)$. The problem is, I don't know how to differentiate that when I need to. So I'd be grateful for some help if someone can think of a different way or just a way to improve mine (using the Cauchy method somehow so that it works).

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    What's that in the exponent of the numerators? – 2017-01-20
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    I've got my glasses on @SachpazisStelios and I believe it is a $4$ :) – 2017-01-20
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    @Mitch Thanks, I finally saw it from editing. – 2017-01-20
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    Sorry, I haven't been here for long and I still had to figure out how to make everything look bigger, haha. Thanks for the edit! – 2017-01-20
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    @Manuel It's fine.:) – 2017-01-20
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    @SimpleArt Have you tried the ratio test? – 2017-01-20
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    @Dr.MV Oh, whoops, was thinking more along $4^n$ rather than $n^4$ :P – 2017-01-20
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    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: [How do I accept an answer?](http://meta.math.stackexchange.com/questions/3286/), [Why should we accept answers?](http://meta.math.stackexchange.com/questions/3399/). – 2017-01-23

3 Answers 3

11

Without Stirling.

Note that the denominator is $$ \sum_{k=1}^n \log k \leq \sum_{k=1}^n \log n = n\log n $$ from which you can lower bound the general term of your series by $$a_n\stackrel{\rm def}{=} \frac{n^4}{\log 1+\log 2+\dots+\log n} \geq \frac{n^3}{\log n}\xrightarrow[n\to\infty]{} \infty$$ and therefore the series $\sum_n a_n$ diverges, as its general term does not even go to $0$.

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    Indeed. You beat me to it by a handful of seconds, though. – 2017-01-20
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    So (+1) for the common approach. – 2017-01-20
8

It is easy to show that $n!\le n^n$.

Therefore, $\log(n!)\le n\log(n)$ and we have

$$\sum_{n=2}^N \frac{n^4}{\log(n!)}\ge \sum_{n=2}^N \frac{n^3}{\log(n)}\to \infty \,\,\text{as}\,\,N\to \infty$$

4

We can use the comparison test to show it diverges and we will be using the series $b_n = n^2$.

First note that $\log(n!) \approx n\log(n) - n$ by the Stirling's formula.

Then $a_n = \frac{n^4}{\log(n!)} \to \frac{n^4}{n\log(n) - n} = \frac{n^3}{\log(n) - 1} > \frac{n^3}{n} = n^2$ which diverges. Then, your series must also diverge.

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    I hadn't thought about using the Stirling's formula! Thank you so much :) – 2017-01-20
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    Stirling's Formula is a bit overkill. Note that $n!\le n^n$ suffices. – 2017-01-20
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    @Dr.MV Sometimes I forget to think simple! Thanks for the good tip :) – 2017-01-20
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    @RSerrao No worry. When I see a factorial, my first thought is often "Stirling." – 2017-01-20