The formula for a geometric series as I know it is $\sum ar^{n-1}$, where $r$ is the common ratio and $a$ is the first number the common ratio is multiplied with.

If we're to conform to that formula, then the series $\sum 1/3^n$ would be $\sum \frac13 \frac{1}{3^{n-1}}$, which would tell you that $a$ = $r$ =$1/3$. From there, using the formula for the sum of a geometric series, $a/1-r$, gives you $1/2$.

However, the sum is apparently equal to $3/2$, which the formula for the sum would only give if you took $a$ to be $1$ instead of $1/3$.

So why is $a$ apparently $1$, despite the fact that I conformed to the formula properly?

Any help is appreciated.

  • $\begingroup$ As Peter Foreman has said in another comment, $a$ in that formula refers to the initial term. So you should think of the $a/(1-r)$ formula as "$\frac{\text{initial term}}{1-\text{common ratio}}$". $\endgroup$ – Minus One-Twelfth Apr 17 at 19:46

Depends if you start the index at $0$ or $1$

  • 1
    $\begingroup$ Wow that makes sense, thank you! So if under the summation was $n=0$, the sum is $3/2$, but if it was $n=1$ it would be $1/2$? Also, does the formula for geometric series become $ar^n$ if the index starts at $0$? $\endgroup$ – James Ronald Apr 17 at 19:30
  • $\begingroup$ The initial term is $a=ar^0$ so when the index is $0$ we get the initial term. $\endgroup$ – Peter Foreman Apr 17 at 19:38

$$ \frac{1}{1-x}=\sum_{k=0}^{\infty}x^k$$ $$\frac{x}{1-x}=\sum_{k=1}^{\infty}x^k$$


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