Integral Inequality involving a binomial

Question

Let $\binom{t}{k}$ where $k\in \mathbb{N}$ be defined as $\frac{t(t-1)\cdots (t-k+1)}{k!}$. Prove that $\int_n^{\infty} \binom{t-1}{n-1} e^{-t} dt \le \frac{1}{(e-1)^n}$

My progress is as follows: we can find a recursion. Let $F(n)$ be the sum in question, then

$$F(n)=\int_n^{n+1} \binom{t-1}{n-1} e^{-t} dt + \int_{n+1}^{\infty} \binom{t-1}{n-1} e^{-t} dt$$

Using Pascal's identity $\binom{t-1}{n-1} = \binom{t-2}{n-1} + \binom{t-2}{n-2}$, one may obtain the recursion that

$$F(n)=\int_n^{n+1} \binom{t-1}{n-1} e^{-t} dt + \frac{F(n)+F(n-1)}{e}$$

$$\frac{e-1}{e} F(n)=\int_n^{n+1} \binom{t-1}{n-1} e^{-t} dt + \frac{F(n-1)}{e}$$

$$F(n)=\int_n^{n+1} \frac{e}{e-1}\binom{t-1}{n-1} e^{-t} dt + \frac{F(n-1)}{e-1}$$

Let $K(n)=F(n)(e-1)^n$ then $K(n)-K(n-1)=e(e-1)^{n-1} \int_n^{n+1} \binom{t-1}{n-1} e^{-t} dt$

So now it makes sense to show $\sum\limits_{n\ge 1} e(e-1)^{n-1} \int_n^{n+1} \binom{t-1}{n-1} e^{-t} dt$ converges to a constant at most 1.

For convenience, we can write it as $$e\sum\limits_{n\ge 0} (e-1)^n \int_{n+1}^{n+2} \binom{t-1}{n} e^{-t} dt = \sum\limits_{n\ge 0} (e-1)^n \int_{n}^{n+1} \binom{t}{n} e^{-t} dt$$

Now I am stuck. If anyone has any ideas on a solution, please don't hesitate to share with us here!

Answer 1

Proceeding along the OP's idea:

Let $$G(n) := (\mathrm{e} - 1)^n\int_n^{\infty} \binom{t-1}{n-1}\mathrm{e}^{-t}\mathrm{d} t.$$

We have $$G(n) = \sum_{k=1}^n \mathrm{e}(\mathrm{e}-1)^{k-1}\int_k^{k+1} \left(\binom{t-1}{k-1} - \binom{t-2}{k-2}\right)\mathrm{e}^{-t}\mathrm{d} t. \tag{1}$$ (The proof is given at the end.)

Using $\binom{t-1}{k-1} - \binom{t-2}{k-2} \ge 0$ for all $k\ge 1$ and $t\in [k, k+1]$, we have \begin{align*} G(n) &= \sum_{k=1}^n \mathrm{e}(\mathrm{e}-1)^{k-1}\int_k^{k+1} \left(\binom{t-1}{k-1} - \binom{t-2}{k-2}\right)\mathrm{e}^{-t}\mathrm{d} t\\ &\le \sum_{k=1}^\infty \mathrm{e}(\mathrm{e}-1)^{k-1}\int_k^{k+1} \left(\binom{t-1}{k-1} - \binom{t-2}{k-2}\right)\mathrm{e}^{-t}\mathrm{d} t\\ &= \sum_{k=1}^\infty \mathrm{e}(\mathrm{e}-1)^{k-1}\mathrm{e}^{-k}\int_0^{1} \left(\binom{t + k-1}{k-1} - \binom{t+k-2}{k-2}\right)\mathrm{e}^{-t}\mathrm{d} t\\ &= \int_0^1 \left[\sum_{k=1}^\infty \mathrm{e}(\mathrm{e}-1)^{k-1}\mathrm{e}^{-k}\left(\binom{t + k-1}{k-1} - \binom{t+k-2}{k-2}\right)\right]\mathrm{e}^{-t}\,\mathrm{d} t\\ &= \int_0^1 \mathrm{e}^{t}\cdot \mathrm{e}^{-t}\mathrm{d} t\\ &= 1 \end{align*} where we use $$\sum_{k=1}^\infty \mathrm{e}(\mathrm{e}-1)^{k-1}\mathrm{e}^{-k}\left(\binom{t + k-1}{k-1} - \binom{t+k-2}{k-2}\right) = \mathrm{e}^{t}. \tag{2}$$ (The proof of (2) is given at the end.)

We are done.

$\phantom{2}$

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Proof of (1):

Let $$F(n) := \int_n^{\infty} \binom{t-1}{n-1}\mathrm{e}^{-t}\mathrm{d} t.$$

Using the identity $\binom{t-1}{n-1} = \binom{t-2}{n-1} + \binom{t-2}{n-2}$, we have \begin{align*} F(n) &= \int_n^{n+1} \binom{t-1}{n-1}\mathrm{e}^{-t}\mathrm{d} t + \int_{n+1}^\infty \binom{t-1}{n-1}\mathrm{e}^{-t}\mathrm{d} t\\[5pt] &= \int_n^{n+1} \binom{t-1}{n-1}\mathrm{e}^{-t}\mathrm{d} t + \int_{n+1}^\infty \binom{t-2}{n-1}\mathrm{e}^{-t}\mathrm{d} t + \int_{n+1}^\infty \binom{t-2}{n-2}\mathrm{e}^{-t}\mathrm{d} t\\[5pt] &= \int_n^{n+1} \binom{t-1}{n-1}\mathrm{e}^{-t}\mathrm{d} t + \mathrm{e}^{-1}\int_{n}^\infty \binom{t-1}{n-1}\mathrm{e}^{-t}\mathrm{d} t + \mathrm{e}^{-1}\int_{n-1}^\infty \binom{t-1}{n-2}\mathrm{e}^{-t}\mathrm{d} t \\ &\qquad - \mathrm{e}^{-1}\int_{n-1}^n \binom{t-1}{n-2}\mathrm{e}^{-t}\mathrm{d} t\\ &= \mathrm{e}^{-1}F(n) + \mathrm{e}^{-1}F(n-1) + \int_n^{n+1} \binom{t-1}{n-1}\mathrm{e}^{-t}\mathrm{d} t - \mathrm{e}^{-1}\int_{n-1}^n \binom{t-1}{n-2}\mathrm{e}^{-t}\mathrm{d} t. \end{align*}

Thus, we have $$F(n) = \frac{1}{\mathrm{e} - 1}F(n-1) + \frac{\mathrm{e}}{\mathrm{e} - 1}\int_n^{n+1} \binom{t-1}{n-1}\mathrm{e}^{-t}\mathrm{d} t - \frac{1}{\mathrm{e} - 1}\int_{n-1}^n \binom{t-1}{n-2}\mathrm{e}^{-t}\mathrm{d} t$$ and $$G(n) = G(n-1) + \mathrm{e}(\mathrm{e}-1)^{n-1}\int_n^{n+1} \binom{t-1}{n-1}\mathrm{e}^{-t}\mathrm{d} t - (\mathrm{e}-1)^{n-1}\int_{n-1}^n \binom{t-1}{n-2}\mathrm{e}^{-t}\mathrm{d} t.$$

Thus, we have \begin{align*} G(n) &= \sum_{k=1}^n \mathrm{e}(\mathrm{e}-1)^{k-1}\int_k^{k+1} \binom{t-1}{k-1}\mathrm{e}^{-t}\mathrm{d} t - \sum_{k=1}^n (\mathrm{e}-1)^{k-1}\int_{k-1}^k \binom{t-1}{k-2}\mathrm{e}^{-t}\mathrm{d} t\\ &= \sum_{k=1}^n \mathrm{e}(\mathrm{e}-1)^{k-1}\int_k^{k+1} \binom{t-1}{k-1}\mathrm{e}^{-t}\mathrm{d} t - \sum_{k=1}^n \mathrm{e}(\mathrm{e}-1)^{k-1}\int_{k}^{k+1} \binom{t-2}{k-2}\mathrm{e}^{-t}\mathrm{d} t\\ &= \sum_{k=1}^n \mathrm{e}(\mathrm{e}-1)^{k-1}\int_k^{k+1} \left(\binom{t-1}{k-1} - \binom{t-2}{k-2}\right)\mathrm{e}^{-t}\mathrm{d} t. \end{align*}

We are done.

$\phantom{2}$

Proof of (2):

Denote $a = \mathrm{e}^{-1}$. Let $$g(k) := \binom{t+k-2}{k-2}.$$

We have \begin{align*} \mathrm{LHS} &= \sum_{k=1}^\infty (1 - a)^{k-1}[g(k + 1) - g(k)]\\ &= -g(1) + \sum_{k=1}^\infty (1 - a)^{k-1} a\, g(k + 1)\\ &= \mathrm{e}^{-1}\sum_{k=0}^\infty (1 - \mathrm{e}^{-1})^k \binom{t + k}{k}\\ &= \mathrm{e}^{-1}\sum_{k=0}^\infty (1 - \mathrm{e}^{-1})^k \binom{-t-1}{k}(-1)^k\\ &= \mathrm{e}^{t}. \end{align*} where we have used $\binom{t+k}{k} = \binom{-t-1}{k}(-1)^k$ and $(1 + x)^n = \sum_{r=0}^\infty \binom{n}{r}x^r$ (the generalized binomial theorem).

We are done.

Answer 2

Let $$ a_n = \int_{n}^{+\infty}\binom{t-1}{n-1}e^{-t}\,dt = e^{-n}\int_{0}^{+\infty}\binom{n-1+t}{n-1}e^{-t}\,dt $$ and $$ A(z) = \sum_{n\geq 1} a_n z^n = \int_{0}^{+\infty}\frac{z}{(e-z)^{t+1}}\,dt = \frac{z}{(e-z)\log(e-z)}=\frac{1}{\log(e-z)}\sum_{m\geq 1}\frac{z^m}{e^m}$$ is an analytic function with radius of convergence given by $e-1$. Once you prove the following

Lemma. For any $n\in\mathbb{N}^+$ the coefficient of $z^n$ in the Maclaurin series of $\frac{1-\frac{z}{e-1}}{\log(e-z)}$ is negative.

^{Hint: https://en.wikipedia.org/wiki/Gregory_coefficients}

you have that

$$ a_n = [z^n] A(z) \leq [z^n]\left(\sum_{r\geq 0}\frac{z^r}{(e-1)^r}\sum_{m\geq 1}\frac{z^m}{e^m}\right)=\sum_{k=0}^{n-1}\frac{1}{(e-1)^ke^{n-k}}=\frac{1-\left(1-\frac{1}{e}\right)^n}{(e-1)^{n-1}}.$$