Show that $\sup_{t \in \mathbb{R}} \sum_{j=1}^\infty \frac{(it)^j}{(k+j)!}$ is bounded for any fixed $k>2$ We already know that $e^{it}$ is bounded by 1 uniformly for all real $t$. And the taylor series of it is actually $\sum_{j=1}^\infty \frac{(it)^j}{(k+j)!}-1$ with $k=0$. Here I figured with $k>2$, the norm is supposed to be smaller. But it is much trickier than I thought. I can't use triangle inequality here since I need a uniform bound over all t. And I can't just factor out something to extract a $e^{it}$ factor. After much work, I haven't been able to show it is uniformly bounded for all real t. Any help will be appreciated.
 A: I guess the fastest way to estimate the function is to observe
\begin{align}
\sum^\infty_{j=1} \frac{(it)^j}{(j+k)!}= e^{it}\frac{(it)^{-k}}{\Gamma(k)}\gamma(k, it) -1=(it)e^{it}\frac{(it)^{-k-1}}{\Gamma(k+1)}\gamma(k+1, it)
\end{align}
where $\Gamma(k)$ is the gamma function and $\gamma(s, z)$ is the lower incomplete gamma function defined by
\begin{align}
\gamma(s, z) = z^s \sum^\infty_{j=0} \frac{(-1)^j z^j}{j! (j+s)}.
\end{align}
Hence it suffices to consider the normalized lower incomplete gamma function
\begin{align}
\left|\frac{(it)^{-k}\gamma(k, it)}{\Gamma(k)} \right|
\end{align}
for $t$ large, say $|t|\gg k$, and show it's bounded. By the series expansion of the normalised incomplete gamma function, we see that
\begin{align}
(it)^{-k}\frac{\gamma(k, it)}{\Gamma(k)} = \frac{1}{\Gamma(k)} \int^\infty_0 e^{-(k-1)s}e^{-it e^{-s}}\ ds.
\end{align}
Next, apply the method of non-stationary phase, we see that
\begin{align}
\left|\frac{1}{\Gamma(k)} \int^\infty_0 e^{-(k-1)s}e^{-it e^{-s}}\ ds \right|\leq \frac{C}{\Gamma(k)|t|}
\end{align}
for some constant $C=C(k)$, can depend on $k$. Hence it follows
\begin{align}
\left|\sum^\infty_{j=1} \frac{(it)^j}{(j+k)!} \right|\leq \frac{C}{ \Gamma(k+1)}.
\end{align}
Note: In fact, one could probably check that $C(k) = 1$. I believe this is as sharp as it will ever get. 
Additional: Observe
\begin{align}
\int^\infty_0 e^{-(k-1)s}e^{-it e^{-s}}\ ds =&\ \frac{1}{it}\int^\infty_0 e^{-(k-2)s}\frac{d}{ds}e^{-it e^{-s}}\ ds\\
=&\ \frac{1}{it}\left[e^{-(k-2)s-ite^{-s}}\bigg|^\infty_{s=0}+(k-2)\int^\infty_0 e^{-(k-2)s-ite^{-s}}\ ds \right]\\
=&\ \frac{e^{-it}}{it}+ \frac{k-2}{it}\int^\infty_0e^{-(k-2)s-ite^{-s}}\ ds.
\end{align} 
Next, note that
\begin{align}
\left|\frac{e^{-it}}{it}+ \frac{k-2}{it}\int^\infty_0e^{-(k-2)s-ite^{-s}}\ ds\right| \leq \frac{1}{|t|}+\frac{k-2}{|t|}\int^\infty_0 e^{-(k-2)s}\ ds = \frac{2}{|t|}.
\end{align}
Hence, we in fact have
\begin{align}
\left|\int^\infty_0 e^{-(k-1)s}e^{-it e^{-s}}\ ds\right| \leq \frac{2}{\Gamma(k)|t|}
\end{align}
where $t$ is big.
