# Does $\sum\ln(n) / n^3$ converge?

By Direct Comparison Test:

$$\ln(n)/n^3 < n/n^3 = 1/n^2$$

Converges as that is a convergent a p-series

By n'th term Test:

$$(\ln(n))' = 1/n$$

$$(n^3)' = 3n^2$$

$$\lim_{n\to\infty} (1/n) / 3n^2 = \lim_{n\to\infty} (3n^2)/n =\lim_{n\to\infty} 3n = \infty$$

so by $n$'th term it diverges.

What am I doing wrong?

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$\frac{a}{b}\neq a*b$. –  rschwieb Aug 16 '12 at 16:23

## 4 Answers

You made an algebraic mistake. You have the quotient $$1/n\over 3n^2.$$ If you wish to "move the numerator $1/n$ downstairs", you need to take its reciprocal, ${1\over 1/n}=n$, first and do it: $${\color{maroon}{1/n}\over 3n^2}={1\over \color{maroon}n\cdot 3n^2}.$$ You did this correctly.

However, when you "moved the denominator $3n^2$ upstairs", you failed to use its reciprocal $1/3n^2$. Done correctly, you would have obtained $${1/n\over\color{maroon}{ 3n^2}}={(1/n)\cdot(\color{maroon}{1/3n^2})\over1 }.$$

Note both methods give $1\over 3n^3$ as a result; there is no need to do both. Perhaps the simplest thing to do in order to simplify your expression is to write $${1/n\over 3n^2}={1\over n}\cdot{1\over 3n^2}={1\over n\cdot3n^2}={1\over 3n^3}.$$

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You have a "typo". The terms do go to zero: $\displaystyle \lim_{n\to \infty}\frac{(1/n)}{3n^2}=\lim_{n\to\infty}\frac{1}{3n^3}=0$.

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You got the division wrong. $$(1/n)/3n^2 = 1/3n^3 \rightarrow 0$$

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You're nth term test is wrong. The reason why you're getting a divergent limit is because you flipped your fractions around inccorectlly. Anyways, the nth term test dosen't involve taking derivitives. But, here is how you can do it:

$\frac{ln(n)}{n^3}>0$ for sufficently large n (i.e n>3). Now, we have that $0<\frac{ln(n)}{n^3}<\frac{1}{n^2}$. Clearly, $\frac{1}{n^2}\rightarrow 0$. Hence, by the squeeze theorem for limits, $\frac{ln(n)}{n^3}$ $\rightarrow 0$. Also, always remember that the nth term test is not sufficent to prove convergance, it's just a nessacary condition. i.e the harmonic series.

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He's taking derivatives in line with L'Hopital's rule. –  process91 Aug 16 '12 at 16:50