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

Why does the following hold:


Can we generalize the above to

n=0xn=11x ?

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

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


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 110.7? (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 Nth partial sum and multiply by 0.7, you get
so that
Solving for sN gives
What is the limit as N? The only part of the expression that depends on N is (0.7)N+1. Since |0.7|<1, then limN(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 r1, 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
S = \frac{a(1-r^{k+1})}{1-r}.

A series of the form


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

= \lim_{N\to\infty}\sum_{n=0}^{N}a
= \lim_{N\to\infty}(N+1)a
= \infty,

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

\sum_{n=0}^{\infty}ar^n = \lim_{N\to\infty}\sum_{n=0}^{N}ar^{N} = \lim_{N\to\infty}\frac{a(1-r^{N+1})}{1-r}.

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

\lim_{N\to\infty}r^{N+1} = \left\{\begin{array}{ll}
0 &\mbox{if $|r|\lt 1$;}\\
1 & \mbox{if $r=1$;}\\
\text{does not exist} &\mbox{if $r=-1$ or $|r|\gt 1$}

it follows that:

\sum_{n=0}^{\infty}ar^{n} &=\left\{\begin{array}{ll}
0 &\mbox{if $a=0$;}\\
\text{diverges}&\mbox{if $a\neq 0$ and $r=1$;}\\
\lim\limits_{N\to\infty}\frac{a(1-r^{N+1})}{1-r} &\mbox{if $r\neq 1$;}\end{array}\right.\\
&= \left\{\begin{array}{ll}
\text{diverges}&\mbox{if $a\neq 0$ and $r=1$;}\\
\text{diverges}&\mbox{if $a\neq 0$, and $r=-1$ or $|r|\gt 1$;}\\
\frac{a(1-0)}{1-r}&\mbox{if $|r|\lt 1$;}
\text{diverges}&\mbox{if $a\neq 0$ and $|r|\geq 1$;}\\
\frac{a}{1-r}&\mbox{if $|r|\lt 1$.}

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

\lim_{N\to\infty}r^{N+1} = \left\{\begin{array}{ll}
0 &\mbox{if $|r|\lt 1$;}\\
1 & \mbox{if $r=1$;}\\
\text{does not exist} &\mbox{if $r=-1$ or $|r|\gt 1$}

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,
|\rho-r^n| = r^n-\rho \leq r^N-\rho \lt\epsilon.
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
|r^{2n}-\rho^2| = |r^n-\rho||r^n+\rho| = (r^n-\rho)(r^n+\rho)\lt (r^n-\rho)(\rho+\epsilon) \lt r^n-\rho\lt\epsilon.
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,
\lim_{N\to\infty}r^{N+1} = \left\{\begin{array}{ll}
0 &\mbox{if $|r|\lt 1$;}\\
1 & \mbox{if $r=1$;}\\
\text{does not exist} &\mbox{if $r=-1$ or $|r|\gt 1$}

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
|r^n| = |r|^n = \frac{1}{(1+a)^n} \leq \frac{1}{1+na} \lt \frac{1}{na}.
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.

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