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Consider the geometric series defined by the sequence,

$a_n=\frac 1 {r^n}$, n=0,1,2,...

Then the n-th partial sum $S_n$ is given by

$S_n=\sum_{k=0}^n \frac 1 {r^k}$

I got $\frac {1-(1/r)^n}{1-(1/r)}$, but it was wrong.

But I used the formula I found above to find $S_n=\sum_{k=0}^\infty \frac 1 {r^k}=\lim_{x \to \infty}S_n= \frac 1 {1-(\frac 1 r)}$ and it worked. So I don't know what went wrong.

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There are $n + 1$ terms in your sum, since it's from $0$ to $n$ inclusive, with the last one being $\frac{1}{r^{n}}$. Thus, such as shown in Geometric series, the formula uses the power of the common ratio, i.e., $\frac{1}{r}$ here, which is one higher than the last term uses, so the sum is actually

$$\frac{1 - \left(\frac{1}{r}\right)^{n+1}}{1 - \left(\frac{1}{r}\right)} \tag{1}\label{eq1A}$$

You can easily verify this, for example, where $n = 0$, then you have that

$$S_0 = \sum_{k=0}^{0}\frac{1}{r^k} = \frac{1}{r^{0}} = 1 \tag{2}\label{eq2A}$$

and \eqref{eq1A} also gives a value of $\dfrac{1-\left(\frac{1}{r}\right)}{1-\left(\frac{1}{r}\right)} = 1$.

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$\sum_{n=0}^k x^n=\dfrac{1-x^{k+1}}{1-x}$. Now, if $|x|\lt1$, then $x^{k+1}\to0$, so we get $\sum_{n=0}^{\infty}x^n=\dfrac1{1-x}$. Thus your answer came out right.

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