# Values of ∑∞n=0xn\sum_{n=0}^\infty x^n and ∑Nn=0xn\sum_{n=0}^N x^n

Why does the following hold:

$$∞∑n=00.7n=11−0.7=10/3?\begin{equation*} \displaystyle \sum\limits_{n=0}^{\infty} 0.7^n=\frac{1}{1-0.7} = 10/3\quad ? \end{equation*}$$

Can we generalize the above to

$$∞∑n=0xn=11−x\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$$ ?

Are there some values of $$xx$$ for which the above formula is invalid?

What about if we take only a finite number of terms? Is there a simpler formula?

$$N∑n=0xn\displaystyle \sum_{n=0}^{N} x^n$$

Is there a name for such a sequence?

This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions.

and here: List of abstract duplicates.

By definition, a “series” (an “infinite sum”)

is defined to be a limit, namely

That is, the “infinite sum” is the limit of the “partial sums”, if this limit exists. If the limit exists, equal to some number $S$, we say the series “converges” to the limit, and we write

If the limit does not exist, we say the series diverges and is not equal to any number.

So writing that

means that we are asserting that

So what your question is really asking is: why is this limit equal to $\frac{1}{1-0.7}$? (Or rather, that is the only way to make sense of the question).

In order to figure out the limit, it is useful (but not strictly necessary) to have a formula for the partial sums,

This is where the formulas others have given come in. If you take the $N$th partial sum and multiply by $0.7$, you get

so that

Solving for $s_N$ gives

What is the limit as $N\to\infty$? The only part of the expression that depends on $N$ is $(0.7)^{N+1}$. Since $|0.7|\lt 1$, then $\lim\limits_{N\to\infty}(0.7)^{N+1} = 0$. So,

Since the limit exists, then we write

More generally, a sum of the form

with $a$ and $r$ constant is said to be a “geometric series” with initial term $a$ and common ratio $r$. If $a=0$, then the sum is equal to $0$. If $r=1$, then the sum is equal to $(k+1)a$. If $r\neq 1$, then we can proceed as above. Letting

we have that

so that

Dividing through by $1-r$ (which is not zero since $r\neq 1$), we get

A series of the form

with $a$ and $r$ constants is called an infinite geometric series.
If $r=1$, then

so the series diverges. If $r\neq 1$, then using the formula above we have:

The limit exists if and only if $\lim\limits_{N\to\infty}r^{N+1}$ exists. Since

it follows that:

Your particular example has $a=1$ and $r=0.7$.

Since this recently came up (09/29/2011), let’s provide a formal proof that

If $r\gt 1$, then write $r=1+k$, with $k\gt0$. By the binomial theorem, $r^n = (1+k)^n \gt 1+nk$, so it suffices to show that for every real number $M$ there exists $n\in\mathbb{N}$ such that $nk\gt M$. This is equivalent to asking for a natural number $n$ such that $n\gt \frac{M}{k}$, and this holds by the Archimedean property; hence if $r\gt 1$, then $\lim\limits_{n\to\infty}r^n$ does not exist. From this it follows that if $r\lt -1$ then the limit also does not exist: given any $M$, there exists $n$ such that $r^{2n}\gt M$ and $r^{2n+1}\lt M$, so $\lim\limits_{n\to\infty}r^n$ does not exist if $r\lt -1$.

If $r=-1$, then for every real number $L$ either $|L-1|\gt \frac{1}{2}$ or $|L+1|\gt \frac{1}{2}$. Thus, for every $L$ and for every $M$ there exists $n\gt M$ such that $|L-r^n|\gt \frac{1}{2}$ proving the limit cannot equal $L$; thus, the limit does not exist. If $r=1$, then $r^n=1$ for all $n$, so for every $\epsilon\gt 0$ we can take $N=1$, and for all $n\geq N$ we have $|r^n-1|\lt\epsilon$, hence $\lim\limits_{N\to\infty}1^n = 1$. Similarly, if $r=0$, then $\lim\limits_{n\to\infty}r^n = 0$ by taking $N=1$ for any $\epsilon\gt 0$.

Next, assume that $0\lt r\lt 1$. Then the sequence $\{r^n\}_{n=1}^{\infty}$ is strictly decreasing and bounded below by $0$: we have $0\lt r \lt 1$, so multiplying by $r\gt 0$ we get $0\lt r^2 \lt r$. Assuming $0\lt r^{k+1}\lt r^k$, multiplying through by $r$ we get $0\lt r^{k+2}\lt r^{k+1}$, so by induction we have that $0\lt r^{n+1}\lt r^n$ for every $n$.

Since the sequence is bounded below, let $\rho\geq 0$ be the infimum of $\{r^n\}_{n=1}^{\infty}$. Then $\lim\limits_{n\to\infty}r^n =\rho$: indeed, let $\epsilon\gt 0$. By the definition of infimum, there exists $N$ such that $\rho\leq r^N\lt \rho+\epsilon$; hence for all $n\geq N$,

Hence $\lim\limits_{n\to\infty}r^n = \rho$.

In particular, $\lim\limits_{n\to\infty}r^{2n} = \rho$, since $\{r^{2n}\}_{n=1}^{\infty}$ is a subsequence of the converging sequence $\{r^n\}_{n=1}^{\infty}$. On the other hand, I claim that $\lim\limits_{n\to\infty}r^{2n} = \rho^2$: indeed, let $\epsilon\gt 0$. Then there exists $N$ such that for all $n\geq N$, $r^n - \rho\lt\epsilon$. Moreover, we can assume that $\epsilon$ is small enough so that $\rho+\epsilon\lt 1$. Then

Thus, $\lim\limits_{n\to\infty}r^{2n} = \rho^2$. Since a sequence can have only one limit, and the sequence of $r^{2n}$ converges to both $\rho$ and $\rho^2$, then $\rho=\rho^2$. Hence $\rho=0$ or $\rho=1$. But $\rho=\mathrm{inf}\{r^n\mid n\in\mathbb{N}\} \leq r \lt 1$. Hence $\rho=0$.

Thus, if $0\lt r\lt 1$, then $\lim\limits_{n\to\infty}r^n = 0$.

Finally, if $-1\lt r\lt 0$, then $0\lt |r|\lt 1$. Let $\epsilon\gt 0$. Then there exists $N$ such that for all $n\geq N$ we have $|r^n| = ||r|^n|\lt\epsilon$, since $\lim\limits_{n\to\infty}|r|^n = 0$. Thus, for all $\epsilon\gt 0$ there exists $N$ such that for all $n\geq N$, $| r^n-0|\lt\epsilon$. This proves that $\lim\limits_{n\to\infty}r^n = 0$, as desired.

In summary,

The argument suggested by Srivatsan Narayanan in the comments to deal with the case $0\lt|r|\lt 1$ is less clumsy than mine above: there exists $a\gt 0$ such that $|r|=\frac{1}{1+a}$. Then we can use the binomial theorem as above to get that

By the Archimedean Property, for every $\epsilon\gt 0$ there exists $N\in\mathbb{N}$ such that $Na\gt \frac{1}{\epsilon}$, and hence for all $n\geq N$, $\frac{1}{na}\leq \frac{1}{Na} \lt\epsilon$. This proves that $\lim\limits_{n\to\infty}|r|^n = 0$ when $0\lt|r|\lt 1$, without having to invoke the infimum property explicitly.