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In which interval (domain) does the sum

$$\sum_{n=1}^{\infty}\log^n(1+x)$$

converge absolutely?

I'm finding difficulty, but if I put $n=1$, then $\log(1+x) \le x$ and $\sum_{n=1}^{\infty}\log^n(1+x)\le \sum x$ is divergent, definitely so I can say that $1$ will not be included in the interval (domain)...

Any hints/solutions will be appreciated.

Thank you

amWhy
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jasmine
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1 Answers1

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HINT

For $\log(1+x)$ to be well defined, $1+x>0$, so $x>-1$.

For $\log^2(1+x)=\log(\log(1+x))$ to be well defined, $\log(1+x)>0$, so $x>0$.

For $\log^3(1+x)$ to be well deefined, $\log^2(1+x)>0$, so $\log(1+x)>1$, so $x>e-1$.

For $\log^4(1+x)$ to be well deefined, $\log^3(1+x)>0$, so $\log^2(1+x)>1$, so $\log(1+x)>e$ so $x>e^e-1$.

Can you follow by induction?

I guess (I didn't do it) that for any $n\in{\mathbb N}$, $\log^n(1+x)$ is well defined iff $x>Exp^{n-2}(e)-1$ (where, $Exp^0(x)=1$ and $Exp^{-1}(x)=0$ and $Exp(x)=e^x$).

Tito Eliatron
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