# Asymptotic behavior of $(1/2 + 2/3 + 3/4 + 4/5 + \cdots + (n-1)/n ) \times n$

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)?

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## migrated from mathematica.stackexchange.comNov 17 '12 at 23:19

This question came from our site for users of Mathematica.

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, you could try using Wolfram|Alpha which is built using Mathematica and is quite capable of evaluating such a simple sum as yours. – Sasha Nov 17 '12 at 23:16

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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Great idea, it also gives 2nd interpretation as n-polygamma(0, n+1)-gamma – Vitaliy Kaurov Nov 17 '12 at 23:48
BTW, this is the URL – Vitaliy Kaurov Nov 18 '12 at 4:42

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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