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I am interested in the following questions:

given:

$G(n) = \left(\frac12 + \frac23 + \frac34 + \frac45 + \cdots + \frac{n-1}n\right)n$

  1. what is a $F(n)$ which could be an upper bound (clearly as tight as possible) for $G(n)$ for $n$ arbitrarily large ?

  2. Does the series: $\frac12 + \frac23 + \frac34 + \frac45 + \cdots + \frac{n-1}n$ have a "name" and a sum (any reference)?

  • 0
    This question most likely does not belong here, but should be migrated to math.SE. This site is specifically about software _Mathematica_, even though _Mathematica_ can do symbolic mathematics. If you have access to _Mathematica_, you can just type your query in and it will provide you with the answer. If you do not have access to [_Mathematica_](http://www.wolfram.com/mathematica), you could try using [Wolfram|Alpha](http://www.wolframalpha.com) which is built using _Mathematica_ and is quite capable of evaluating such a simple sum as yours.2012-11-17

4 Answers 4

8

Let $F(n) = \left(\dfrac{2-1}{2} + \dfrac{3-1}{3} + \dfrac{4-1}{4} + \cdots + \dfrac{n-1}{n} \right) $ we then have that $F(n) = \left(1 - \dfrac12 + 1 - \dfrac13 + 1 - \dfrac14 + \cdots + 1 - \dfrac1n \right) = (n-1) - \left( \dfrac12 + \dfrac13 + \cdots + \dfrac1n \right)$ Now note that $-\left( \dfrac12 + \dfrac13 + \cdots + \dfrac1n \right) = 1 - H_n = 1 - \left(\log(n) + \gamma + \dfrac1{2n} - \dfrac1{12n^3} + \mathcal{O}(1/n^5) \right)$ Hence, $F(n) = n - \left(\log(n) + \gamma + \dfrac1{2n} - \dfrac1{12n^3} + \mathcal{O}(1/n^5) \right)$ $nF(n) = n^2 - n \log n -\gamma n - \dfrac12 + \dfrac1{12n^2} + \mathcal{O}(1/n^4)$ You could make use of the fact that $H_n = \log (n) + \gamma - \dfrac{\zeta(0)}n + \sum_{k=1}^{\infty} \dfrac{\zeta(-k)}{n^{k+1}}$ to get better approximations/bounds.

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Clearly

$\begin{align*} \frac12+\frac23+\frac34+\frac45+\ldots+\frac{n-1}n&=n-1-\sum_{k=2}^n\frac1k\\ &=n-\sum_{k=1}^n\frac1k\\ &=n-H_n\;, \end{align*}$

where $H_n$ is the $n$-th harmonic number, so $G(n)=n^2-nH_n\;.$

There are good approximations of $H_n$ by relatively nice functions:

$H_n\sim\ln n+\gamma+\frac1{2n}-\frac1{12n^2}\;,$

for instance, and the approximation can be improved by taking more terms of the series

$H_n\sim\ln n+\gamma+\frac1{2n}-\sum_{k\ge 1}\frac{B_{2k}}{2kn^{2k}}\;,$

where the numbers $B_k$ are the Bernoulli numbers and $\gamma$ is the Euler-Mascheroni constant.

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Your expression in the parentheses is $(n-1)-\left(\frac12+\frac13+\cdots+\frac{1}{n}\right)$ which is asymptotically related to $n-\ln n$. So in total you have something like $n^2-n \ln n. $Does that help?

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Regarding your second question: Just type "sum 1/2 + 2/3 + 3/4 + 4/5 + 5/6 + ... + (n - 1)/n" or something similar in WolframAlpha. According to WolframAlpha this sum does not have a name but is equal to $n - {H_n}$ where ${H_n}$ is the $n$-th harmonic number. Source: WolframAlpha.

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    BTW, [**this is the URL**](http://www.wol$f$ramalpha.com/input/?i=sum+1%2F2+%2B+2%2F3+%2B+3%2F4+%2B+4%2F5+%2B+5%2F6+%2B+...+%2B+%28n+-+1%29%2Fn)2012-11-18