Calculating the sum of this arithmetic series: $9-6+4- \frac 83 + ... + \frac{256}{729}-\frac{512}{2187}$

Question

I need help with calculating the sum of this arithmetic series:
$9-6+4- \frac 83 + ... + \frac{256}{729}-\frac{512}{2187}$

I watched this math video to try to solve it: https://youtu.be/BA0uxIaMtMs
And I found these two formulas:
$a_n=a_1 + (n-1)d$
$S_n= \frac n2 (a_1+a_n)$
But what about the subtractions? And this arithmetic series is also shrinking since $9 \gt \frac{256}{729}$, how do I handle that? I did however find that the right side of the subtraction is always $\frac 23$ of left side, but I don't know how to use that info. Any help is appreciated.

Answer 1

Arithmetic progression is sequence $a_n$ such that difference of consecutive terms is constant. That is for any $n, m$, $ a_{n + 1} - a_n = a_{m + 1} - a_m$. Formulas you gave are true for any arithmetic progressions.

But your progression isn't arithmetic. For example $(-6) - 9 \neq 4 - (-6)$. Sequence obeying rule $a_{n+1} = q \cdot a_n$ for some constant $q$, we call geometric progression. In this case, the quotient of consecutive terms is constant (and equal to $q$). (last part doesn't work for almost zero sequences)

If you start numbering your sequence at $0$, you can easily prove that $a_n = a_0 \cdot q^n$. To figure out partial sum of geometric series you can use formula $$ \sum_{k = 0}^n a_k = \sum_{k = 0}^n a_0 q^k = a_0(1 + q + q^2 + \ldots + q^n) = a_0\frac{1 - q^{n+1}}{1 - q} $$ whenever $q \neq 1$. On the other hand, when $q = 1$ the sequence is constant, and you can easily compute the sum.

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In your case you have $q = - \frac 2 3$, $a_0 = 9$ and $n=9$.

Answer 2

HINT:

Note that $a_n=(-2/3)a_{n-1}$ and $a_0=9$. Write down the expression for the general term $a_n$.

What type of series is $\sum_{n=0}^\infty a_n$? It is not an arithmetic series.