# How prove this inequality $\sum\limits_{k=0}^{n}\frac{\sum\limits_{i=k}^{n}\binom{i}{k}}{k+1}\cdot\left(-\frac{1}{e}\right)^{k+1}<1$

show that $$\sum_{k=0}^{n}\left(\dfrac{\displaystyle\sum_{i=k}^{n}\binom{i}{k}}{k+1}\cdot\left(-\dfrac{1}{e}\right)^{k+1}\right)<1$$

maybe this inequality is from Mathematical olympiad,

I think $\dfrac{\sum_{i=k}^{n}\binom{i}{k}}{k+1}$can use integral define?

and maybe use Abel transformation, $$\sum_{k=1}^{n}a_{k}b_{k}=S_{n}b_{n}+\sum_{k=1}^{n-1}S_{k}(b_{k}-b_{k+1})$$ where $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$

so let $$a_{k}=\left(-\dfrac{1}{e}\right)^{k+1},b_{k}=\dfrac{\displaystyle\sum_{i=k}^{n}\binom{i}{k}}{k+1}\Longrightarrow S_{n}=\dfrac{1}{e^2+e}\left(1-\left(-\dfrac{1}{e}\right)^n\right)$$ so use Abel transformation, $$\sum_{k=0}^{n}\left(\dfrac{\displaystyle\sum_{i=k}^{n}\binom{i}{k}}{k+1}\cdot\left(-\dfrac{1}{e}\right)^{k+1}\right)=\dfrac{1}{e^2+e}\left(1-\left(-\dfrac{1}{e}\right)^n\right)\cdot\dfrac{1}{n+1}+\sum_{k=1}^{n-1}\dfrac{1}{e^2+e}\left(1-\left(-\dfrac{1}{e}\right)^k\right)\left[\dfrac{\binom{k}{k}+\binom{k+1}{k}+\cdots+\binom{n}{k}}{k+1}-\dfrac{\binom{k+1}{k+1}+\binom{k+2}{k+1}+\cdots+\binom{n}{n}}{k+2}\right]$$ maybe someone can solve it? Thank you

• I was unable to completely answer the question but it may help someone else to notice that: $$\sum_{k=0}^{n}\bigg(\sum_{j=k}^{n}\binom{j}{k}\frac{(\frac{-1}{e})^{k+1}}{k+1}\bigg)=-\sum_{j=0}^{n}\bigg(\sum_{k=0}^{j}\binom{j}{k}\int_{\frac{-1}{e}}^{0}x^{k}dx\bigg)$$ $$=-\int_{\frac{-1}{e}}^{0}\sum_{j=0}^{n}(1+x)^{j}dx=-\int_{\frac{-1}{e}}^{0} \frac{(1+x)^{n+1}-1}{x}dx$$ $$=\sum_{k=1}^{n+1}\binom{n+1}{k}\frac{(\frac{-1}{e})^{k}}{k}$$ – user71352 Jul 7 '14 at 17:01

It can be found that \begin{align} \sum_{i=k}^{n} \binom{i}{k} = \binom{n+1}{n-k} \end{align} for which \begin{align} S_{n} = \sum_{k=0}^{n} \frac{(-e^{-1})^{k+1}}{k+1} \left(\sum_{i=k}^{n} \binom{i}{k}\right) \end{align} becomes \begin{align} S_{n} = (n+1) \sum_{k=0}^{n} \binom{n}{k} \frac{(-e^{-1})^{k+1}}{(k+1)^{2}}. \end{align} By using the relation \begin{align} \sum_{k=0}^{n} \binom{n}{k} \frac{t^{k}}{(k+1)^{2}} = {}_{3}F_{2}(-n, 1, 1; 2, 2; -t) \end{align} it is seen that \begin{align} S_{n} = - \frac{1}{e} \ {}_{3}F_{2}(-n, 1, 1; 2, 2; e^{-1}). \end{align} It is evident that $S_{n} < 0$ and thus \begin{align} \sum_{k=0}^{n} \frac{(-e^{-1})^{k+1}}{k+1} \left(\sum_{i=k}^{n} \binom{i}{k}\right) < 1. \end{align}