Could someone please explain how to work out the limit of :
$$\frac{n\log(n)}{(n+1)\log(n+1)},\qquad\mbox{ as $n\to\infty$}.$$
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1The expression in your title and in your question are different. Which is it? Also, please make it clear which terms are in the numerator and which are in the denominator. – John Jan 29 '14 at 22:14
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1$\lim\limits_{n\to\infty} \frac{n}{n+1} = 1$ $\implies$ $\lim\limits_{n\to\infty} \frac{\log(n)}{\log(n+1)} = 1$ $\implies$ $...$ – Oleg567 Jan 29 '14 at 22:14
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You lost a $+1$ in the argument of the second logarithm compared to your title. Also, please use parentheses to show that the second log is in the denominator (isn't it?) – Ross Millikan Jan 29 '14 at 22:14
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Ah sorry, it should be how it is stated in the title, the denominator should read : (n+1)log(n+1). Sorry! – Sadie Jan 29 '14 at 22:16
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@Oleg567 That needs a bit more justification; after all, $\lim\frac{n}{n+1}=1\not!!!\implies \lim\frac{2^n}{2^{n+1}}=1$... – Steven Stadnicki Jan 29 '14 at 22:21
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@StevenStadnicki, yes, indeed. I was too quick with (draft) comment. – Oleg567 Jan 29 '14 at 22:30
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Thank you everyone for your help, I was trying to use the ration test to show the sum of 1/(nlogn) diverges but since the sum appears to equal 1 the ratio test is inconclusive. Can anyone suggest another way of showing that the sum of 1/(nlogn) from 2 to infinity diverges please? – Sadie Jan 29 '14 at 22:40
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@Sadie Try the integral test; $\int\frac1{x\log x}dx$ can be evaluated explicitly (hint: try substituting $u=\log x$) and that should be enough to give you your answer. – Steven Stadnicki Jan 29 '14 at 23:06
2 Answers
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$$\ln(n+1)=\ln n\bigg(1+\frac1n\bigg)=\ln n+\ln\bigg(1+\frac1n\bigg)$$
$$\lim_{n\to\infty}\frac{n\ln n}{(n+1)\ln(n+1)}=\lim_{n\to\infty}\frac n{n+1}\cdot\lim_{n\to\infty}\frac{\ln n}{\ln(n+1)}=1\cdot\lim_{n\to\infty}\frac{\ln n}{\ln n+\ln\bigg(1+\frac1n\bigg)}=$$
$$=\lim_{n\to\infty}\frac1{1+\dfrac{\ln\bigg(1+\frac1n\bigg)}{\ln n}}=\frac1{1+0}=1.$$
Lucian
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How do you say that $\dfrac{\ln(1+\frac{1}{n}}{\ln n}) =0$ as $n\to \infty$ – Learnmore Jun 17 '16 at 05:44
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@learnmore: $~\dfrac1n\to0~$ and $~\ln1=0~$ and $~\dfrac1{\ln n}\to0~$ as $~n\to\infty.$ – Lucian Jun 17 '16 at 11:16
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Turn to functions on $\mathbb{R}_{>0}$ and apply L'Hospital $\left(\frac{\infty}{\infty}\right)$ twice, if you know about it (if not, somebody else will attempt an elementary proof for you):
$$ \lim_{x\to\infty} \frac{x\ln x }{(x+1)\ln(x+1)} = \lim_{x\to\infty} \frac{\ln x +1 }{\ln(x+1) +1} = \lim_{x\to\infty} \frac{1/x }{1/(x+1)} = 1$$
ir7
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