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I'm having difficulty with this problem on my homework.

"Find the value of x such that series $(\ln x)^n$ converges."

I've tried taking the natural log, so it would be $n \ln(\ln x)$ then $(\ln(\ln x))/(1/n)$ for lhospitals rule. But I'm not fairing to well, anyone have any hints on what to do or things that I might be doing wrong with this approach?

Thanks.

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    $\begingroup$ Let $y = \ln x$. Can you figure out for which values of $y$ the series $y^n$ converges? $\endgroup$ – FredH Mar 25 '19 at 18:43
  • $\begingroup$ If y = ln x, then ln x would have to be -1 < r < 1 as its geometric, thats genius, thanks! $\endgroup$ – Christian Martinez Mar 25 '19 at 18:44
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We essentially are asking for which $x$ does

$$\sum_{k=0}^\infty \ln(x)^n = \sum_{k=0}^\infty r^n$$

converge, where $r = \ln(x)$. With this in mind, we visibly have a geometric series of ratio $r = \ln(x)$. Thus, since we require $|r| < 1$ for such series to converge, we need $x$ such that

$$|\ln(x)| < 1 \iff -1 < \ln(x) < 1 \iff \frac 1 e < x < e$$

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    $\begingroup$ Do you mean $\frac1e\lt x\lt e$? $\endgroup$ – Peter Foreman Mar 25 '19 at 19:22
  • $\begingroup$ Yeah, I derped. Thanks! $\endgroup$ – Eevee Trainer Mar 25 '19 at 22:25

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