Question: Find the value of $\sum_{i=1}^{n}(\frac{1}{n-i})^{c}$ for large $n$.

\begin{align} \sum_{i=1}^{n}(\frac{1}{n-i})^{c} & = \sum_{i=1}^{n}(\frac{1}{n})^{c}(\frac{1}{1-\frac{i}{n}})^{c} \\ & = \frac{n}{n} \times \sum_{i=1}^{n}(\frac{1}{n})^{c}(\frac{1}{1-\frac{i}{n}})^{c} \\ & = n(\frac{1}{n})^{c} \sum_{i=1}^{n}\frac{1}{n}(\frac{1}{1-\frac{i}{n}})^{c} \qquad(1) \end{align}

Let $f(x) = (\frac{1}{1-x})^{c}$, by using Riemann-sum theorem, we have \begin{align} \lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac{1}{n}(\frac{1}{1-\frac{i}{n}})^{c} & = \int_{0}^{1} (\frac{1}{1-x})^{c} = A \qquad(2) \end{align} By using $(1)$ and $(2)$, for sufficently large $n$, we have $$\bbox[5px,border:2px solid #C0A000]{\sum_{i=1}^{n}(\frac{1}{n-i})^{c} = A\times n(\frac{1}{n})^{c}}$$

The presented proof has a problem, $f(x)$ is not defined in the closed interval $[0,1]$. How can I solve this problem?

Definition (Riemann-sum theorem) Let $f(x)$ be a function dened on a closed interval $[a, b]$. Then, we have $$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}f\Big(a +(\frac{b - a}{n})i\Big)\frac{1}{n}=\int_{a}^{b}f(x)dx$$

  • $\begingroup$ The problem is that the question doesn't ask for the limit as $n\to\infty$ $\endgroup$ – saulspatz Jun 9 '18 at 15:52
  • $\begingroup$ Since the sum contains a term $\bigl(\frac{1}{0}\bigr)^c$, it may be a trick question. Or it may be that $\sum_{i = 0}^{n-1}$ was intended. $\endgroup$ – Daniel Fischer Jun 9 '18 at 17:45

\begin{align}\label{eq:7777} & \frac{2}{\sqrt{n-i} + \sqrt{n-i+1}} \leq \frac{1}{\sqrt{n-i}} \leq \frac{2}{\sqrt{n-i} + \sqrt{n-i-1}} \nonumber\\ & \qquad \Rightarrow 2(\sqrt{n-i+1} - \sqrt{n-i}) \leq \frac{1}{\sqrt{n-i}} \leq 2(\sqrt{n-i} - \sqrt{n-i-1}) \nonumber\\ & \qquad \Rightarrow 2 \sum_{i=1}^{n-1}(\sqrt{n-i+1} - \sqrt{n-i}) \leq \sum_{i=1}^{n-1} \frac{1}{\sqrt{n-i}} \leq 2 \sum_{i=1}^{n-1}(\sqrt{n-i} - \sqrt{n-i-1}) \nonumber\\ & \qquad \Rightarrow 2 (\sqrt{n}-1) \leq \sum_{i=1}^{n-1} \frac{1}{\sqrt{n-i}} \leq 2 \sqrt{n-1} \end{align}


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