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I wish to find a closed form for $\sum_{i=1}^n\frac{1}{i}$. does it exist? If so, what is it?

I cannot arrive at one using any methods I am aware of.

bob the pie
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    It's called the $n$-th harmonic number, $\sum_{i = 1}^n \frac{1}{i} = H_n$. Not really a closed form, but the asymptotics are known pretty well. – Daniel Fischer Jul 30 '15 at 20:43
  • Any approximations? – bob the pie Jul 30 '15 at 20:45
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    $H_n = \log \bigl(n+\frac{1}{2}\bigr) + \gamma + O(n^{-2})$, where $\gamma$ is the Euler-Mascheroni constant. – Daniel Fischer Jul 30 '15 at 20:47
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    As $n$ becomes large, your limit becomes $\ln n + \gamma$ where $\gamma$ is the Euler constant, and $\ln n$ is natural logarithm. I don't know if it's proved that no closed form exists, but at least, no simple closed form is known. – user2566092 Jul 30 '15 at 20:47
  • DOES this have a closed form and no one has found it yet, or is it known that it does not have a closed form solution? – Race Bannon Jul 30 '15 at 21:31
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    @RaceBannon Notion of "closed form" is highly informal one, and there is no standard definition for that. This makes it impossible to argue that it does not have a closed form. – Wojowu Jul 30 '15 at 21:49
  • @wojowu what about if we call closed form to an expression for which the amount of operations to compute the result doesn't depend of the argument of the expression? – YoTengoUnLCD Aug 06 '15 at 02:19
  • @YoTengoUnLCD So $n!$ has no closed form? – Matt Samuel Aug 10 '15 at 22:33

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The following could be called a closed form in absence of a better answer. Define $$e_k(x_1,x_2,\ldots,x_n)=\sum_{1\leq i_1<i_2<\cdots<i_k\leq n}{x_{i_1}x_{i_2}\cdots x_{i_k}}$$ This is the $k$th elementary symmetric polynomial in $x_1,x_2,\ldots,x_n$. Then $$\sum_{i=1}^n{\frac{1}{i}}=\frac{e_{n-1}(1,2,3,\ldots,n)}{n!}$$ We can see this by taking a common denominator. Of course, the denominator $n!$ will not generally be as small as possible. For example, $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{1\cdot 2\cdot 3+1\cdot 2\cdot 4+1\cdot 3\cdot 4+2\cdot 3\cdot 4}{1\cdot 2\cdot 3\cdot 4}=\frac{50}{24}=\frac{25}{12}$$

Matt Samuel
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