Proving a sum of three series is equal to $e^z$ Let
\begin{align*}
    u(z)=&\ 1+\frac{z^3}{3!}+\frac{z^6}{6!}+\frac{z^9}{9!}+\ldots\\
    v(z)=&\ z+\frac{z^4}{4!}+\frac{z^7}{7!}+\frac{z^{10}}{10!}+\ldots\\
    w(z)=&\ \frac{z^2}{2!}+\frac{z^5}{5!}+\frac{z^8}{8!}+\frac{z^{11}}{11!}+\ldots
\end{align*}
show that $e^z= u(z)+ v(z) +w(z) $
We can rewrite the three series as
$$u(z)=\sum^{\infty}_{n=0}\frac{z^{3n}}{(3n)!}$$
$$v(z)=\sum^{\infty}_{n=0}\frac{z^{3n+1}}{(3n+1)!}$$
$$w(z)=\sum^{\infty}_{n=0}\frac{z^{3n+2}}{(3n+2)!}$$
Thus
$$ u(z)+ v(z) +w(z) = \sum^{\infty}_{n=0} z^{3n} \left(\frac{1}{(3n)!}+\frac{z}{(3n+1)!}+\frac{z^2}{(3n+2)!}\right)$$
but I cant really see how to combine the factorial terms together. Any ideas where I should go from here? I suppose I could just claim that the sum of all the terms appears to form the sequence $\frac{z^n}{n!}$, but that doesn't seem rigorous enough to me.
 A: Any non-negative integer $n$ is of the form $3k+i$ for some non-negative integer $k$ and some $i \in \{0,1,2\}$. Check that the coefficient of $z^{n}=z^{3k+i}$ in $u(z)+v(z)+w(z)$ is exactly $\frac 1 {(3k+i)!}=\frac 1 {n!}$ in each of the cases $i=0, i=1$ and $i=2$.
A: 
I suppose I could just claim that the sum of all the terms appears to form the sequence $\frac{z^n}{n!}$, but that doesn't seem rigorous enough to me.

Well, you could make it more rigorous by appealing to the known calculus result which is:

If $\sum_{i=0}^\infty a_nx^n$ and $\sum_{i=0}^\infty b_nx^n$ are both convergent series, then $\sum_{i=0}^\infty (a_n + b_n)x^n$ is also a convergent sequence and $\sum_{i=0}^\infty a_nx^n+\sum_{i=0}^\infty b_nx^n=\sum_{i=0}^\infty (a_n+b_n)x^n$

A: It suffices to show that for each complex $z \in \mathbb{C}$ the exponential series can be brought to the same form that you obtained in the right-hand side of your final relation. The most elementary method to achieve this is to realise that:
$$\mathrm{e}^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}=\lim_{n \to \infty}\sum_{k=0}^n \frac{z^k}{k!}=\lim_{n \to \infty}\sum_{k=0}^{3n+2}\frac{z^k}{k!},$$
bearing in mind that in general if a complex sequence $u \in \mathbb{C}^{\mathbb{N}}$ has limit $a \in \mathbb{C}$ then the subsequence $(u_{kn+l})_{n \in \mathbb{N}}$ will have the same limit $a$ for any $k \in \mathbb{N}^*=\mathbb{N}\setminus\{0\}$ and $l \in \mathbb{N}$ (in our case $k=3$ and $l=2$).
Finally, since the map:
$$\begin{align*}
\left(\mathbb{N} \cap [0, n]\right) \times \{0, 1, 2\} &\to \mathbb{N} \cap [0, 3n+2] \\
(h, l) &\mapsto 3h+l
\end{align*}$$
is a bijection
it is clear that we can obtain the arrangement:
$$\sum_{k=0}^{3n+2}\frac{z^k}{k!}=\sum_{h=0}^n\left(\frac{z^{3h}}{(3h)!}+\frac{z^{3h+1}}{(3h+1)!}+\frac{z^{3h+2}}{(3h+2)!}\right),$$
which is precisely the general term of the sequence whose limit you are considering in the aforementioned right-hand side.
A: Let $U_n(z),V_n(z),W_n(z)$ be the partial sums of the $n$ first terms. As all three series are convergent,
$$U_\infty(z)+V_\infty(z)+W_\infty(z)=\lim_{n\to\infty}U_n(z)+\lim_{n\to\infty}V_n(z)+\lim_{n\to\infty}W_n(z)
\\=\lim_{n\to\infty}(U_n(z)+V_n(z)+W_n(z))=\lim_{n\to\infty}E_{3n}(z)=E_\infty(z)=e^z$$
where $E_m(z)$ denotes the sum of the $m$ first terms of the exponential development.

Technical note:
$E_{3n}(z)$ indeed converges to $e^z$ as it is a subsquence of the converging $E_m(z)$.
