# Sequences and Series: Find the value for x such that series (ln x)^n is convergent

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.

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

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
$$|\ln(x)| < 1 \iff -1 < \ln(x) < 1 \iff \frac 1 e < x < e$$
• Do you mean $\frac1e\lt x\lt e$? – Peter Foreman Mar 25 '19 at 19:22