Taylor expansion of $e^t$ starting with $N\geq0$ real number

Question

For each $N\in[0,\infty)$ define $$ f_N(t)=\sum_{i=0}^\infty \dfrac{t^{N+i}}{\Gamma(N+1+i)}=\dfrac{t^N}{\Gamma(N+1)}+\dfrac{t^{N+1}}{\Gamma(N+2)}+\cdots,\quad t\geq0. $$ It is like the Taylor expansion of $e^t$ but starting with a real number $N$ and since $N!$ is not defined we consider $\Gamma(N+1)$ instead of $N!$, where $\Gamma$ is the Gamma function. I am trying to prove that $$ f_N(t)\leq e^t,\quad\forall t\in[0,\infty). $$ I think that holds the following more general inequality: $f_N(t)\leq f_M(t)$ if $M\leq N$. Using WolframAlpha, this inequality is plausible.

My Conclusions:

Firstly, it is trivial for $N\in\mathbb N$, and to prove the more general inequality, we can assume $N-M\leq1$. Moreover, the case $t\leq1$ follows by $$ f_N(t)\leq \sum_{i=0}^\infty\dfrac{t^{\lceil N-1+i\rceil}}{\lceil N-1+i\rceil!}\leq e^t, $$ where $\lceil\cdot\rceil$ is the ceiling function.

My try 1: Trying to prove $t^{N+i}/\Gamma(N+1+i)\leq t^{\lceil N+i\rceil}/\lceil N+i\rceil!$ or $t^{N+i}/\Gamma(N+1+i)\leq t^{\lceil N-1+i\rceil}/\lceil N-1+i\rceil!$.

My try 2: Finding the Taylor expansion of $f_N$ at $t=1$ and compare with $e^t$.

My try 3: Define $g_N(t):=e^t-f_N(t)$ and try to prove that $g_N(t)>0$. Since $g_N(0)=1$ and (by the graph in WolframAlpha) $g'_N(t)>0$ we could conclude $g_N(t)>0$. But I do not know how to prove $g'_N(t)>0$.

My try 4: Define $h_N(t):=f_N(t)/e^t$ and try to prove that $h_N(t)\leq1$. Since $h_N(0)=0$ and (by WolframAlpha) $h_N'(t)>0$ we could conclude $h_N(t)\leq1$ since $\lim_{t\to\infty}h_N(t)=1$. I do not know how to prove $h'_N(t)>0$ and $\lim_{t\to\infty}h_N(t)=1$.

Answer 1

Firstly, let us prove the equality given by @River Li \begin{equation} f_N(t)=e^t-e^t\dfrac{N}{\Gamma(N+1)}\Gamma(N,t).\qquad\qquad\qquad(1) \end{equation} For $N=0$ is trivial. If $N>0$ we can define $g_N\colon [0,\infty)\to\mathbb R$ such that $$ g_N(t)=f_N(t)-e^t+e^t\dfrac{N}{\Gamma(N+1)}\Gamma(N,t). $$ Note that $$ g_N(0)=-1+\dfrac{N}{\Gamma(N+1)}\Gamma(N)=0 $$ and \begin{align*} g_N'(t)&=\dfrac{N}{\Gamma(N+1)}t^{N-1}+f_N(t)-e^t+\dfrac{N}{\Gamma(N+1)}\left[e^t\int_t^\infty s^{N-1}e^{-s}\mathrm ds-e^tt^{N-1}e^{-t}\right]\\ &=f_N(t)-e^t+e^t\dfrac{N}{\Gamma(N+1)}\int_t^\infty s^{N-1}e^{-s}\mathrm ds\\ &=g_N(t). \end{align*} Hence $g_N\equiv0$. This concludes the equation (1).

Now given $1\leq M<N$ define $h_{N,M}\colon(0,\infty)\to\mathbb R$ by $$ h_{N,M}(t)=\dfrac{f_N(t)}{f_M(t)}. $$ We have $h_{N,M}(t)\leq1$ as an immediate consequence of

1. $\lim_{t\to0}h_{N,M}(t)=0$;
2. $\lim_{t\to\infty}h_{N,M}(t)=1$;
3. $h_{N,M}$ is increasing.

To deal with 1. we note that $$ h_{N,M}(t)=\dfrac{\frac{t^{N}}{\Gamma(N+1)}+\frac{t^{N+1}}{\Gamma(N+2)}+\cdots}{\frac{t^{M}}{\Gamma(M+1)}+\frac{t^{M+1}}{\Gamma(M+2)}+\cdots}=\dfrac{\frac{t^{N-M}}{\Gamma(N+1)}+\frac{t^{N-M+1}}{\Gamma(N+2)}+\cdots}{\frac{1}{\Gamma(M+1)}+\frac{t}{\Gamma(M+2)}+\cdots}\overset{t\to0}\longrightarrow 0. $$ We see that $$ \lim_{t\to\infty}h_{N,M}(t)=\lim_{t\to\infty}\dfrac{1-\dfrac{N}{\Gamma(N+1)}\int_t^\infty s^{N-1}e^{-s}\mathrm ds}{1-\dfrac{M}{\Gamma(M+1)}\int_t^\infty s^{M-1}e^{-s}\mathrm ds}=1, $$ which is clear from (1). Now we are left with the task to check $h_{N,M}'(t)>0$. Using $\Gamma(N+1+j)=\Gamma(N+1)\prod_{i=0}^{j-1}(N+1+i)$ we have \begin{align*} h'_{N,M}(t)f_M(t)^2&=\dfrac{N}{\Gamma(N+1)}t^{N-1}f_{M+1}(t)-\dfrac{M}{\Gamma(M+1)}t^{M-1}f_{N+1}(t)\\ &=\sum_{j=0}^\infty\left(\dfrac{N}{\Gamma(N+1)\Gamma(M+1+j)}-\dfrac{M}{\Gamma(M+1)\Gamma(N+1+j)}\right)t^{N+M-1}\\ &=\sum_{j=0}^\infty\dfrac{N\prod_{i=0}^{j-1}(N+1+i)-M\prod_{i=0}^{j-1}(M+1+i)}{\Gamma(M+1+j)\Gamma(N+1+j)}t^{N+M-1}\\ &>0. \end{align*}