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$$\lim_{n\to \infty}(n-n^2\cdot ln(1+\frac{1}{n}))$$

What I tried doing was the following:

$$=\lim_{n\to \infty}n(1-n\cdot ln(1+\frac{1}{n}))=\lim_{n\to \infty}n(1- ln(1+\frac{1}{n})^n)=\lim_{n\to \infty}n(1-ln(e))=$$ $$=\lim_{n\to \infty}n(1-1)=\infty \cdot 0$$

And to be honest, I have no idea on how to proceed or what to do. The result is supposed to be $\frac{1}{2}$ according to Wolfram Alpha, but it's not capable of showing the steps to the solution.

Thanks for the help.

Jonathan Martinez
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  • $\lim_{n\to\infty}n(1- ln(1+\frac{1}{n})^n) = \lim_{n\to\infty}n(1-\ln(e))$ is wrong. – Martín-Blas Pérez Pinilla Jan 26 '17 at 17:04
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    Time to pass to more serious tools: $\log(1+x)=x-\frac12x^2+o(x^2)$ when $x\to0$ hence $n-n^2\ln\left(1+\frac1n\right)=n-n^2\left(\frac1n-\frac12\frac1{n^2}+o\left(\frac1{n^2}\right)\right)=\frac12+o(1)\to\frac12$. – Did Jan 26 '17 at 17:04
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    $3$rd question of http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion – lab bhattacharjee Jan 26 '17 at 17:13
  • @Did Ok thanks, but is it valid to use Mcloren even though it's a sequence and not a function? – Jonathan Martinez Jan 27 '17 at 09:07
  • Of course, Taylor expansions are legitimate here. Say, what are your sources on this? (Dunno what is "Mcloren" though.) – Did Jan 27 '17 at 13:31
  • It's a taylor sequence when evaluated at x=0 or something.. the source is a test from my college – Jonathan Martinez Jan 27 '17 at 14:15

2 Answers2

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So have $$\lim\limits_{x\to 0}\frac{\ln (1+x)-x}{x^2}=-\frac{1}{2}$$ which follows from l'Hospitals rule or other methods.

Setting $x=\frac{1}{n}$ we get your limit.

Rene Schipperus
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Using Jack D'Aurizio method::

$$\lim_{n\rightarrow \infty}\bigg(n-n^2 \ln\left(\frac{n+1}{n}\right)\bigg) = \lim_{n\rightarrow \infty}n\int^{n+1}_{n}\left(\frac{x-n}{x}\right)dx$$

Put $x-n=t\;,$ Then $dx=dt$ and changing limits, We get

$$ = \lim_{n\rightarrow \infty} n\int^{1}_{0}\frac{t}{n+t}dt =\lim_{n\rightarrow \infty} \int^{1}_{0}\frac{t}{1+\frac{t}{n}}dt =\lim_{n\rightarrow \infty}\int^{1}_{0} tdt = \frac{1}{2}$$

juantheron
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