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I have to check whether series is convergent or not.

$$\sum_{n=1}^{\infty}\frac{\ln(n)}{n(n+1)}$$

I used condensation test for this so i get new series as

$$ \frac{n \ln(2)}{2^n(2^n+1)}.$$ Now I apply ratio test for new series and I get $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{4}$. So my original series is also convergent. However is this correct way, I am not quite sure.

Thanks

davyjones
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Taylor Ted
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    You need to replace $a_n$ with $2^n a_{2^n}$, not $a_{2^n}$. The $2^n$ factor in your denominator will then cancel. However, the ratio test can stil finish the proof; you get $\tfrac{1}{2}$ for the limit rather than $\tfrac{1}{4}$. – J.G. Jan 18 '16 at 08:38

2 Answers2

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You can do a bound test for large $n$.

For $n>N$ where $N$ is some bound, the following is true: $$\frac{\ln n}{n(n+1)}<n^{-2+\epsilon}$$ where $1>\epsilon>0$.

We know that $\sum n^k$ converges for $k<-1$. I used the fact that $\mathcal{O}(\ln n)<\mathcal{O}(n^\epsilon)$ for any $\epsilon>0$.

orion
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  • where can i find inequalities like this used in series – Taylor Ted Jan 18 '16 at 08:42
  • This is one of the two famous rate of growth comparisons: exponential functions grow faster than any power (can be verified by noticing that taylor series has all positive terms and doesn't end), and logarithm, that is slower than any positive power (can be verified by doing the inverse of the above: inverse of $e^x$ is slower than inverse of $x^n$ for arbitrarily big n, so it's slower than $x^{1/n}$ for $n\to\infty$. Otherwise, search for big O notation, you have a lot of comparisons there. – orion Jan 18 '16 at 10:18
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Here's another way to prove it. We know that $\ln n <\sqrt{n}$ starting from some $n$.

Since $\frac{\ln n}{n(n+1)}<\frac{\sqrt{n}}{n(n+1)}=\frac1{\sqrt{n}(n+1)}\sim \frac1{n^{3/2}}$ and we know that series $\sum_{k=1}^\infty \frac1{k^\alpha}$ converges if $\alpha>1$. Therefore $$\sum_{n=1}^\infty \frac{\ln n}{n(n+1)}$$ converges.

Rasmus Erlemann
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