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auto23652im 2022-07-12 Answered
Consider the sum
${S}_{n}=\sum _{r=2}^{n}\frac{\left(-1{\right)}^{r}\left(\genfrac{}{}{0}{}{n}{r}\right)}{r-1}$
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## Answers (1)

karburitc
Answered 2022-07-13 Author has 7 answers
Note that
$\begin{array}{rl}{S}_{n}& =\sum _{r=2}^{n}\frac{\left(-1{\right)}^{r}}{r-1}\left(\genfrac{}{}{0}{}{n}{r}\right)=n\sum _{r=1}^{n-1}\frac{\left(-1{\right)}^{r+1}}{r\left(r+1\right)}\left(\genfrac{}{}{0}{}{n-1}{r}\right)\\ & =n\sum _{r=1}^{n-1}\frac{\left(-1{\right)}^{r+1}}{r}\left(\genfrac{}{}{0}{}{n-1}{r}\right)+n\sum _{r=1}^{n-1}\frac{\left(-1{\right)}^{r}}{r+1}\left(\genfrac{}{}{0}{}{n-1}{r}\right)\\ & =n{H}_{n-1}+n{\int }_{0}^{1}\left[\left(1-x{\right)}^{n-1}-1\right]dx=n{H}_{n-1}-n\frac{n-1}{n}\\ & =n{H}_{n}-n.\end{array}$
Thus, we have an exact expression for ${S}_{n}$ in terms of ${H}_{n}$. The asymptotic expansion follows from that of ${H}_{n}$, i.e.,
${S}_{n}\sim n\mathrm{log}n-\left(1-\gamma \right)n+\frac{1}{2}-\frac{1}{12n}+\frac{1}{120{n}^{3}}-\cdots$
as $n\to +\mathrm{\infty }$

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