# Is the sum of the series $\sum 1/3^n$ equal to $1/2$ or $3/2$?

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.

• 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 "$\frac{\text{initial term}}{1-\text{common ratio}}$". – Minus One-Twelfth Apr 17 at 19:46

## 2 Answers

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

• 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 $0$? – James Ronald Apr 17 at 19:30
• The initial term is $a=ar^0$ so when the index is $0$ we get the initial term. – Peter Foreman Apr 17 at 19:38

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