Following limit appears a bit hard to evaluate to anything reasonable:$\lim_{n\rightarrow\infty}=\left(1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}\right) \ln\frac{2n+1}{n}$umm... any ideas, hints?
Horrible limit, ask for any hints
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calculus
limits
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0@clark: just try it using
. My approaches $1$ as far as I let it run... – 2012-11-07
1 Answers
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It’s much easier than it looks. First,
$\lim_{n\to\infty}\ln\frac{2n+1}n=\lim_{n\to\infty}\left(2+\frac1n\right)=\ln 2\;,$
so you can ignore this factor and concentrate on
$\lim_{n\to\infty}\left(1+\sqrt2+\sqrt[3]3+\ldots+\sqrt[n]n\right)\;.$
Clearly $\sqrt[n]n\ge 1$ for each positive integer $n$, so ... ?
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0@user46034: You’re welcome. Don’t take the comments too personally: those two reactions are something to watch out for, but Harald and I really were talking about a general phenomenon, not about you in particular. And don’t be too sure that you didn’t try hard enough: sometimes you just get caught in a mental blind alley and don’t see a way out no matter how hard you try. (Leaving the problem for a while and coming back to it for a fresh start sometimes helps.) – 2012-11-07