consequence of supremum of a series Given $f: \mathbb N \to \mathbb R$. Asume that 
$$A:=\sup_{k\geqslant 0}\Big| \sum_{n=0}^\infty \frac{t^n}{n!}f(k+n)\Big|<\infty,$$
for some $t\in (0,1)$, can we conclude that $f$ is bounded on $\mathbb N$? My proof as follow:


*

*If $f\geqslant 0$, then $\infty > A \geqslant \sup_{k\geqslant 0}|f(k)|$ (consider at $n=0$).

*Otherwise, we write $f=f^+-f^-$ and can we apply the above to get the conclusion? This is the point that I'm not sure.
 A: We denote by $\ell^{\infty}\left(\mathbb{R}\right)$ the space of bounded sequences in $\mathbb{R}$. That is:
\begin{equation*}
\ell^{\infty}\left(\mathbb{R}\right)=\left\{\left(a_{0},a_{1},\ldots\right)\in\mathbb{R}^{\omega}\bigm\vert\sup_{i\ge0}\left\{\left\lvert a_{i}\right\rvert\right\}<+\infty\right\}.
\end{equation*}
We equip this space with the norm $\left\lvert\left\lvert{}a\right\rvert\right\rvert_{\ell^{\infty}\left(\mathbb{R}\right)}=\sup_{i\ge0}\left\{\left\lvert a_{i}\right\rvert\right\}$.
Define the linear operator $L_{t}:\ell^{\infty}\left(\mathbb{R}\right)\to\ell^{\infty}\left(\mathbb{R}\right)$ by
\begin{equation*}
\left(L_{t}(a)\right)_{i}=\sum_{n=0}^{\infty}\frac{t^{n}}{n!}a_{i+n}
\end{equation*}
for $i\ge0$ and for fixed $t\in\mathbb{R}$. Note that this operator is bounded since
\begin{equation*}
\left\lvert\left(L_{t}(a)\right)_{i}\right\rvert=\left\lvert\sum_{n=0}^{\infty}\frac{t^{n}}{n!}a_{i+n}\right\rvert\le\left\lvert\left\lvert a\right\rvert\right\rvert_{\ell^{\infty}\left(\mathbb{R}\right)}\sum_{n=0}^{\infty}\frac{t^{n}}{n!}=e^{t}\left\lvert\left\lvert a\right\rvert\right\rvert_{\ell^{\infty}\left(\mathbb{R}\right)}.
\end{equation*}
Since $i\ge0$ was arbitrary we obtain that
\begin{equation*}
\left\lvert\left\lvert L_{t}(a)\right\rvert\right\rvert_{\ell^{\infty}\left(\mathbb{R}\right)}\le{}e^{t}\left\lvert\left\lvert a\right\rvert\right\rvert_{\ell^{\infty}\left(\mathbb{R}\right)}.
\end{equation*}
Similarly define $M_{t}:\ell^{\infty}\left(\mathbb{R}\right)\to\ell^{\infty}\left(\mathbb{R}\right)$ by
\begin{equation*}
\left(M_{t}(a)\right)_{i}=\sum_{n=0}^{\infty}\frac{(-t)^{n}}{n!}a_{i+n}
\end{equation*}
for $i\ge0$. Observe that $\left\lvert\left\lvert M_{t}\right\rvert\right\rvert_{\ell^{\infty}\left(\mathbb{R}\right)}\le{}e^{t}\left\lvert\left\lvert a\right\rvert\right\rvert_{\ell^{\infty}\left(\mathbb{R}\right)}$ and that
\begin{align*}
\left(M_{t}\left(L_{t}(a)\right)\right)_{i}&=\sum_{n=0}^{\infty}\frac{(-t)^{n}}{n!}\left(L(a)\right)_{i+n}=\sum_{n=0}^{\infty}\frac{(-t)^{n}}{n!}\sum_{m=0}^{\infty}\frac{t^{m}}{m!}a_{i+n+m}\\
&=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{(-1)^{n}(n+m)!}{n!m!}\frac{t^{n+m}}{(n+m)!}a_{i+n+m}\\
&=\sum_{s=0}^{\infty}\sum_{k=0}^{s}(-1)^{k}\begin{pmatrix}s\\k\\\end{pmatrix}\frac{t^{s}}{s!}a_{i+s}=\sum_{s=0}^{\infty}\frac{t^{s}}{s!}a_{i+s}\left(1+(-1)\right)^{s}=a_{i}.
\end{align*}
Since $i\ge0$ was arbitrary this holds in general. Similarly we also have
\begin{equation*}
\left(L_{t}\left(M_{t}(a)\right)\right)_{i}=a_{i}
\end{equation*}
for $i\ge0$. Thus, $M_{t}$ is the inverse operator to $L_{t}$. We conclude that $L_{t}$ is bijective on $\ell^{\infty}\left(\mathbb{R}\right)$. Thus, the norms $\left\lvert\left\lvert\cdot\right\rvert\right\rvert_{\ell^{\infty}\left(\mathbb{R}\right)}$ and $\left\lvert\left\lvert L_{t}\left(\cdot\right)\right\rvert\right\rvert_{\ell^{\infty}\left(\mathbb{R}\right)}$ are equivalent. Observe that
\begin{equation*}
\left(L_{t}(f(n))\right)_{i}=\sum_{n=0}^{\infty}\frac{t^{n}}{n!}f(i+n)
\end{equation*}
is in $\ell^{\infty}\left(\mathbb{R}\right)$ by the information given. Thus, $(f(n))_{n\ge1}\in\ell^{\infty}\left(\mathbb{R}\right)$.
A: There is another interesting way to prove that $f$ is bounded. First of all an important pre-requisite:

If $g(x)$ is a power series and derivatives of all order of $g$ exist at some $r\in R$, i.e. $g^{(0)}(x), g^{(1)}(x),...$ converge at $r$, then $g(x)=\sum_{n=0}^{\infty}g^{(n)}(r)\frac{(x-r)^n}{n!} \ \forall \ x \in \mathbb R$. The proof is a wonderful exercise in summation notation.
Proof: Say $g(x)=\sum_{n=0}^{\infty}b_nx^n$. Then $$\sum_{n=0}^{\infty}g^{(n)}(r)\frac{(x-r)^n}{n!}=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}\frac{(k+n)!}{k!}b_{k+n}r^k\right)\frac{(x-r)^n}{n!}$$ $$ = \sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{(k+n)!}{(k!)(n!)}b_{k+n}r^{k}\left(\sum_{m=0}^n{n \choose m}(-1)^{(n-m)}r^{(n-m)}x^m\right) $$ $$ \sum_{n=0}^{\infty}\sum_{k=0}^{\infty} \sum_{m=0}^{n} {{k+n}\choose k}{n \choose m}b_{k+n}(-1)^{(n-m)}r^{(k+n-m)}x^m$$You can sum over $m$ in the end, i.e. $$\sum_{m=0}^{\infty}\left(  \sum_{n=m}^{\infty} \sum_{k=0}^{\infty} {{k+n}\choose k}{n \choose m}b_{k+n}(-1)^{(n-m)}r^{(k+n-m)} \right)x^m$$ Now take $j=n+k$ and see that for given $n$, $j$ goes from $n$ to $\infty$, so that the sum becomes $$\sum_{m=0}^{\infty}\left(  \sum_{n=m}^{\infty} \sum_{j=n}^{\infty} {{j}\choose n}{n \choose m}b_{j}(-1)^{(n-m)}r^{(j-m)} \right)x^m $$ Now the final part: we can interchange the two inner summations of $j$ and $n$. Notice that $j\ge n \ge m$ so that the sum becomes 
$$ \sum_{m=0}^{\infty}\left(  \sum_{j=m}^{\infty} \sum_{n=m}^{j} {{j}\choose n}{n \choose m}b_{j}(-1)^{(n-m)}r^{(j-m)} \right)x^m $$ 
$$  \sum_{m=0}^{\infty}\left(  \sum_{j=m}^{\infty} \left( \sum_{n=m}^{j} {{j}\choose n}{n \choose m} (-1)^{(n-m)} \right) b_{j}r^{(j-m)} \right)x^m  $$We have considerably resolved this summation. We only need to show that the terms for $j>m$ vanish. 
To see this notice that $$(1+x)^j=\sum_{p=0}^j{j\choose p}x^{(j-p)}$$ and $$(1+x)^{-(m+1)}=\sum_{q=m}^{\infty}{q\choose m}(-1)^{(q-m)}x^{(q-m)}$$ and that the expression $\sum_{n=m}^{j} {{j}\choose n}{n \choose m} (-1)^{(n-m)}$ is nothing but the coefficient of $x^{(j-m)}$ in $(1+x)^j (1+x)^{-(m+1)}$, i.e. the coefficient of $x^{(j-m)}$ in $(1+x)^{(j-m-1)}$. If $j=m$ this is equal to one. However if $j>m$ then $j-m-1\ge 0$ and thus coefficient is equal to zero. Thus  $$ \sum_{n=m}^{j} {{j}\choose n}{n \choose m} (-1)^{(n-m)} = \delta_{jm}$$. Therefore the summation becomes $$ \sum_{m=0}^{\infty} b_{m}r^{(m-m)} x^m = g(x)$$ as required.

Now consider the power series $h(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}f(n)$. Notice that $$h^{(k)}(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}f(n+k) \ \forall \ k\ge0$$It follows that $h^{(k)}(x)$ converges at $x=t \ \forall \ k\ge0$ and $h^{(k)}(t)=\sum_{n=0}^{\infty}\frac{t^n}{n!}f(n+k)$. Thus $h(x)=\sum_{n=0}^{\infty}h^{(n)}(t)\frac{(x-t)^n}{n!}$. Now notice that $ |h^{(k)}(t) \frac{(x-t)^k}{k!}| < A \frac{|x-t|^k}{k!} \ \forall \ k\ge0$ and since $\sum_{n=0}^{\infty} A\frac{|x-t|^n}{n!}$ converges everywhere in $\mathbb R$, it follows from Weierstrass-M test that the series $\sum_{n=0}^{\infty}h^{(n)}(t)\frac{(x-t)^n}{n!}$ converges everywhere in $\mathbb R$. Thus $h(x)$ converges everywhere in $\mathbb R$. Also $|h(x)| < A \left( \sum_{n=0}^{\infty}\frac{|x-t|^n}{n!} \right)=Ae^{|x-t|}$.
Similarly, by differentiating $h(x)= \sum_{n=0}^{\infty}h^{(n)}(t) \frac{(x-t)^n}{n!}$you can show that $$h^{(k)}(x)=\sum_{n=0}^{\infty}h^{(k+n)}(t)\frac{(x-t)^n}{n!}$$ $\forall \ k \ge 0$. Again it follows from Weierstrass-M test that $h^{(k)}(x)$ converges everywhere in $\mathbb R$ and $$|h^{(k)}(x)|= \left| \sum_{n=0}^{\infty}h^{(k+n)}(t)\frac{(x-t)^n}{n!} \right| < A \left( \sum_{n=0}^{\infty} \frac{|x-t|^n}{n!} \right) = Ae^{|x-t|}$$. Thus 
$$|h(0)|=|f(0)| < Ae^{|0-t)|}=Ae^{t}...$$
$$|h^{(k)}(0)| = |f(k)| < Ae^{|0-t|} = Ae^{t} \ \forall \ k \ge 0$$ proving that $f$ is bounded. 
