Relation between two sequences or summations Let us define two sequences
\begin{equation}
G_{n}=\sum_{i=1}^n a^{-i}t_{i-1}
\end{equation}
and
\begin{equation}
g_{n}=\sum_{i=1}^n t_{i-1}
\end{equation}
where $a$ is an integer and $t_n$ is an another sequence.
Can we find a relation between $G_n$ and $g_n$ ?
 A: Here are  relationships of $G_n$ and $g_n$ in terms of the delta operator $\Delta$ and of generating functions which could be  helpful.
We  define the  generating functions
\begin{align*}
G(z)=\sum_{n\geq 1}G_{n}z^n\qquad\text{and}\qquad
g(z)=\sum_{n\geq 1}g_{n}z^n
\end{align*}
and show

The following is valid
  \begin{align*}
\Delta G_n&=\frac{1}{a^{n+1}}\Delta g_n\qquad\qquad n\geq 1\tag{1}\\
&\\
g(z)&=\frac{1-az}{1-z}G(az)\tag{2}
\end{align*}

$$ $$

Ad (1) We obtain
\begin{align*}
\Delta G_n&=G_{n+1}-G_n\\
&=\sum_{i=1}^{n+1}a^{-i}t_{i-1}-\sum_{i=1}^{n}a^{-i}t_{i-1}\\
&=a^{-(n+1)}t_{n}\\
&=\frac{1}{a^{n+1}}\left(\sum_{i=1}^{n+1}t_{i-1}-\sum_{i=1}^{n}t_{i-1}\right)\\
&=\frac{1}{a^{n+1}}\left(g_{n+1}-g_n\right)\\
&=\frac{1}{a^{n+1}}\Delta g_n\\
&\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
Ad (2) We obtain
\begin{align*}
g(z)&=\sum_{n=1}^\infty g_nz^n=\sum_{n=1}^\infty\left(\sum_{i=1}^{n}t_{i-1}\right)z^n\\
&=\sum_{n=1}^\infty\left(\sum_{i=0}^{n-1}t_{i}\right)z^n=\sum_{n=0}^\infty\left(\sum_{i=0}^{n}t_{i}\right)z^{n+1}\tag{3}\\
&=\frac{z}{1-z}\sum_{n=0}^\infty t_nz^{n}=\frac{z}{1-z}\sum_{n=0}^\infty \frac{t_n}{a^n}\left(az\right)^n\tag{4}\\
&=\frac{z}{1-z}(1-az)\sum_{n=0}^\infty\left(\sum_{i=0}^n\frac{t_i}{a^i}\right)(az)^n\tag{5}\\
&=\frac{z}{az}\frac{1-az}{1-z}\sum_{n=0}^\infty\left(\sum_{i=0}^n\frac{t_i}{a^i}\right)(az)^{n+1}\\
&=\frac{z}{az}\frac{1-az}{1-z}\sum_{n=1}^\infty\left(\sum_{i=0}^{n-1}\frac{t_i}{a^i}\right)(az)^{n}\\
&=\frac{1-az}{1-z}\sum_{n=1}^\infty\left(\sum_{i=1}^{n}\frac{t_{i-1}}{a^i}\right)(az)^{n}\\
&=\frac{1-az}{1-z}G(az)\\
&\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}

Comment:


*

*In (3) we shift the index $i$ by $1$ and then $n$ by $1$.

*In (4) we use
\begin{align*}
\frac{1}{1-z}\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}a_k\right)z^n
\end{align*}

*In (5) we do it similarly as we did it in (4) but the other way round.
