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I try to find the limit of the sequence: $X_n =\displaystyle\sum_{k=1}^n\frac{1}{k} - \ln(n)$.

I try to find it on WolfarmAlpha, but It didn't reconize this limit.

Thank you

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    Do you mean $X_n=\displaystyle\sum_{k=1}^n\frac{1}{k}-\ln n$?2012-01-13
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    @Paul: Yes. Thanks.2012-01-13
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    I have edited your post. Please check if this is what you would like to ask. Thanks.2012-01-13
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    [Harmonic Number](http://mathworld.wolfram.com/HarmonicNumber.html)2012-01-13
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    The limit is called the Euler–Mascheroni constant, but it has no known closed form, and this limit is basically its definition. See: http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant2012-01-13
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    See also proof that [Euler's constant exists](http://planetmath.org/encyclopedia/ProofThatEulersConstantExists.html) at PlanetMath. More generally for any positive function $f(x)$ which decreases to $0$, the limit $\gamma_f=\lim\limits_{n\to\infty} \sum_{k=1}^n f_k-\int_1^n f(x) \mathrm{d}x$ exists. See [Maclaurin-Cauchy Theorem](http://mathworld.wolfram.com/Maclaurin-CauchyTheorem.html) at MathWorld.2012-01-13

2 Answers 2

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Euler's constant is $\lim \limits_{n\to \infty } \displaystyle\sum_{k=1}^n\frac{1}{k} - \ln(n)%$.it is very famous in number theory and analytic number theory.Euler's constant is open problem just we can show this series is converges.

you can look page 52 from Introduction to analytic number theory, Volume 1 By Tom M. Apostol for more detail

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    bagben: It is not clear what you meant by *Euler's constant is open problem*. Perhaps you wanted to say that the question whether this number is rational an open problem?2012-01-13
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    Martin Sleziak: thanks I mean this number is rational an open problem2012-01-13
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If you need to show existence of the limit, show that $X_n$ is decreasing and bounded below.