Relation between two sequences or summations

Question

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 $g_n$ ?

Answer 1

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.