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How do we sum up this series in terms of $n$?

$$1 + \frac{1}{2} + \ldots+\frac{1}{n}$$

Can we create a formula in terms of $n$ for this series?

jimjim
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user674669
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    There is no "formula" for this one as there are for say geometric series or arithmetic series. There are certain approximation formulas which you may study in calculus II. – Maesumi Mar 05 '13 at 18:34
  • Well, Stirling numbers provide an explicit formula of sorts. You can use $H_n\sim\log n+\gamma$ if you want. There are better bounds available on Wikipedia, and since it's more or less the digamma function you could even have a power series expansion. – anon Mar 05 '13 at 18:37
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    You may be interested in this – Maesumi Mar 05 '13 at 18:38
  • I am new to this site. How do I read the answers with all the $$ and \frac. A quick look at the faq didn't help. I am surprised humans can read this - looks like the answers are formatted for computers. – user674669 Mar 05 '13 at 19:26
  • @user674669 : What are you talking about? that is the nice math rendering? what browser are you using? You should see $$ things rendered as nice equations. Let me know if you are not seeing the pretty math – jimjim Mar 05 '13 at 21:20
  • I am seeing the nice expressions on a different computer and browser. – user674669 Mar 07 '13 at 17:20

3 Answers3

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$$\sum_{k=1}^n\frac{1}{k}=\ln n+\gamma+\Theta\big(\frac{1}{2n}\big)$$ where $\gamma $ is Euler–Mascheroni constant.

Aang
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    The $O(\ln n)$ makes the rest of the expression redundant - you mean to say, I think, that it's simply $\ln n + \gamma + O(1/n)$ (or $\Theta(1/n)$ for the last term), or $\ln n+\gamma + \frac{1}{2n} + o(1/n)$ but the current expression is a bit nonsensical. – Steven Stadnicki Mar 05 '13 at 18:54
  • This is true. But it is also true with $\ln n$ instead of $O(\ln n)$. – Julien Mar 05 '13 at 18:56
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As I explained here, for $n>1$,

$$1/1+1/2+\cdots+1/n =\log(n+1/2) + \gamma + \varepsilon(n)$$

with

$$0 < \varepsilon(n) < 1/(24n^2)$$

which gives a good approximation.

Community
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Charles
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Let $\displaystyle H_n=\sum_{k=1}^n\frac{1}{k}$. First, we'll prove that the sequence $(u_n)=(H_n-\log n)$ converges by showing that the series $\displaystyle\sum_n u_{n}-u_{n-1}$ is convergent. We have: $$u_{n}-u_{n-1}=\frac{1}{n}+\log(1-\frac{1}{n})\sim-\frac{1}{2n^2},$$ so the series $\displaystyle\sum_n u_n$ is convergent by comparaison with the Riemann series.

If we denote by $\gamma$ the limit of the sequence $(u_n)$ then we have the formula: $$\sum_{k=1}^n\frac{1}{k}=\log(n)+\gamma+o(1).$$ Moreover, we have: $$\sum_{k=n+1}^{\infty}u_{k}-u_{k-1}=\gamma-u_n\sim-\sum_{k=n+1}^{\infty}\frac{1}{2k^2}\sim-\frac{1}{2n},$$ hence, we have an improved formula: $$\sum_{k=1}^n\frac{1}{k}=\log(n)+\gamma+\frac{1}{2n}+o(\frac{1}{n}).$$