What does this sequence converge to? $a_n = \ln(n+1) - \ln(n)$

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So wolfram has that this limit is 0:

so I think I made a mistake:

$$a_n = \ln(n+1) - \ln(n)$$

$$a_n = \frac{\ln(n+1)}{\ln(n)}$$

using this theorem:

and L'Hospital rule:

$$a_n = \frac{\ln(n+1)}{\ln(n)}$$

$$ \lim_{x\to\infty} = \frac{\ln(x+1)}{\ln(x)}$$

$$ \lim_{x\to\infty} = \frac{\frac{1}{x+1}}{{\frac{1}{x}}}$$

$$ \lim_{x\to\infty} = \frac{(x)}{(x+1)}$$

L'Hospital again:

$$ \lim_{x\to\infty} = \frac{1}{1} = 1$$

What did I do wrong?

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2 Answers

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It is not true that$$\ln(n+1) - \ln(n) = \frac{\ln(n+1)}{\ln(n)}$$

Instead,

$$\ln(n+1)-\ln(n) = \ln\left(\frac{n+1}{n}\right)$$

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This is $$ \int_n^{n+1} \frac{1}{x} dx $$and therefore between $0$ and $\frac{1}{n}$. Another way to see that the limit is 0.

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