$\lim\limits_{n \to \infty} \left( \frac{\ln(n+1)^{n+1}}{\ln n^n} \right)^n$
I've been trying everything for hours with no luck. The result should be $e$. Any good ideas?
How to calculate $\lim\limits_{n \to \infty} \left( \frac{\ln(n+1)^{n+1}}{\ln n^n} \right)^n$?
1
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sequences-and-series
limits
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0Maybe you can substitute, like $lnx=t$? Never mind about $n$ being integer. – 2017-01-21
1 Answers
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Using $ln(a^b) = b*ln(a)$, write the expression as ($\frac{n+1}{n})^n * (\frac{ln(n+1)}{ln(n)})^n$, and go on from there. I guess the problem becomes now the second limit.Using ln(a*b) = ln(a) + ln(b) : $$\lim_{n\to\infty} \Bigg{(}\frac{ln(n+1)}{ln(n)}\Bigg)^n = \lim_{n\to\infty} \Bigg[\Big(1+\frac{ln(n+1)-ln(n)}{ln(n)}\Big)^{\frac{ln(n)}{ln(n+1)-ln(n)}}\Bigg]^{n*\big(\frac{ln(n+1)-ln(n)}{ln(n)}\big)} = e^{\lim_{n\to\infty}n*\big(\frac{ln(n+1)-ln(n)}{ln(n)}\big)} = e^{\lim_{n\to\infty}n*\big(\frac{ln(n+1)}{ln(n)}-1\big)} = e^{\lim_{n\to\infty}n*\big(\frac{ln(n)}{ln(n)}+\frac{ln(1+\frac{1}{n})}{ln(n)}-1\big)} = e^{\lim_{n\to\infty}n*\frac{ln(1+\frac{1}{n})}{ln(n)}} = e^{\lim_{n\to\infty}\frac{\big(ln(1+\frac{1}{n})^n\big)}{ln(n)}} = e^0 = 1$$
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0Won't that give $e * e^{-1}$?... which is not quite right. – 2017-01-21
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0Check again.Are you sure? – 2017-01-21
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0My calculation: $\frac{\ln (n+1)}{\ln n} = \frac{\frac{1}{n+1}}{\frac{1}{n}}=\frac{n}{n+1}=\left(\left( 1- \frac{1}{n+1} \right)^{-(n+1)} \right)^{\frac{n}{-(n+1)}} = e^{-1}$ i ommited the n power until the result of the l'Hospital. – 2017-01-21
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0Seems like you can't use l'Hospital on $\left(\frac{\ln (n+1)}{\ln n}\right)^n$. The result of it is 1 so now the question becomes how to reach it. – 2017-01-21
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0Note that if $a_n\to a$ then $(1+\frac{a_n}{n})^n\to e^a$. – 2017-01-21