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

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 $$\lim_{t\to\infty}h_N(t)=1$$.

• It is enough to consider $1 \le N \le 2$ since $$f_N (t) = f_{N - \left\lfloor N \right\rfloor +1} (t) -\sum\limits_{k=1}^{\left\lfloor N \right\rfloor -1} {\frac{{t^{N - k} }}{{\Gamma (N + 1 - k)}}} .$$ Your function is related to that of Mittag–Leffler since $f_N (t) = t^N E_{1,N + 1} (t)$. From the $t\to +\infty$ asymptotics of that function, $$f_{N - \left\lfloor N \right\rfloor + 1} (t) = e^t-\frac{{t^{N-\left\lfloor N \right\rfloor }}}{{(N - \left\lfloor N \right\rfloor )!}}+\mathcal{O}(t^{N-\left\lfloor N \right\rfloor -1})$$ which confirms the inequality for large $t$ at least.
– Gary
Commented Mar 3, 2022 at 2:29
• Thank you for your comment @Gary. I do not know these Mittag-Leffler function. Would you recommend me a reference? Where can I find one proof of the asymptotic? Commented Mar 3, 2022 at 19:11
• I used Maple to evaluate $f_N(t) = \mathrm{e}^t - \mathrm{e}^t \frac{N}{\Gamma(N + 1)}\Gamma(N, t)$ where $\Gamma(a, x)$ is the upper incomplete gamma function. Commented Mar 4, 2022 at 3:07

Firstly, let us prove the equality given by @River Li $$$$f_N(t)=e^t-e^t\dfrac{N}{\Gamma(N+1)}\Gamma(N,t).\qquad\qquad\qquad(1)$$$$ For $$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 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*}