# How to evaluate series. eg. $a_n = \frac{1}{2^n}$ [duplicate]

Possible Duplicate:
Value of $\sum\limits_n x^n$

In my lecture notes:

$$a_n = \frac{1}{2^n}, \qquad \sum_{n-1}^{\infty} a_n = \lim_{n\to\infty} (1-\frac{1}{2^n}) = 1$$

How do I get $(1-\frac{1}{2^n})$?

A similar example given is

$$a_n = 2^{n-1}, \qquad \sum_{n-1}^{\infty} a_n = \lim_{n\to\infty} (2^n - 1)$$

How do I get these?

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## marked as duplicate by Aryabhata, Zev ChonolesMar 11 '12 at 18:35

Those are from the standard formula for the sum of a finite geometric series:

$$\sum_{k=0}^nar^k=\frac{a-ar^{n+1}}{1-r}=\frac{ar^{n+1}-a}{r-1}\;.\tag{1}$$

Write $$S=\sum_{k=0}^nar^k\;;$$ then

\begin{align*}S&=a\color{red}{+ar+ar^2+\dots+ar^{n-1}+ar^n}\\ rS&=\quad\;\;\,\color{red}{ar+ar^2+\dots+ar^{n-1}+ar^n}+ar^{n+1}\;, \end{align*}

and when you subtract the second equation from the first, the red terms cancel out to leave $(1-r)S=a+ar^{n+1}$, from which $(1)$ follows immediately.

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$$a_n = \frac{1}{2^n}, \qquad \sum_{n=1}^{\infty} a_n = \lim_{n\to\infty} (1-\frac{1}{2^n}) = 1$$

Depends how you interpret it. First let us look at the solution to

First we will be using (for $r<1$)

$$\sum_{k=0}^{n} ar^{k} = \frac{a- ar^{n+1}}{1-r}$$ $$\sum_{k=0}^\infty ar^k = \frac{a}{1-r}$$

Now in the problem above

$$a_n = \frac{1}{2^n}$$ and therefore

\begin{align*} \sum_{n=1}^{\infty} a_n &= \sum_{n=1}^{\infty}\frac{1}{2^n}\\ &= {\frac{1}{2}} ( 1 + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \dots )\\ &= {\frac{1}{2}} (\frac{1}{1-\frac{1}{2}}) &= 1 \end{align*}

And

$$\lim_{n\to\infty} (1-\frac{1}{2^n}) = 1 - \lim_{n\to\infty} \frac{1}{2^n} = 1$$

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It is just a way to rewrite the series which is also explained on the relevant wikipedia article.

Basically all you have to know is that

$$(1+r+r^2+r^3+\ldots+r^n)(1-r)=1-r^{n+1}$$

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