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


*

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

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