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Why does the following hold:

\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

$\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ ?

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?

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

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.

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By definition, a "series" (an "infinite sum") $$\sum_{n=k}^{\infty} a_n$$ is defined to be a limit, namely $$\sum_{n=k}^{\infty} a_n= \lim_{N\to\infty} \sum_{n=k}^N a_n.$$ 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 $$\sum_{n=k}^{\infty} a_n = S.$$ If the limit does not exist, we say the series diverges and is not equal to any number.

So writing that $$\sum_{n=0}^{\infty} 0.7^n = \frac{1}{1-0.7}$$ means that we are asserting that $$\lim_{N\to\infty} \sum_{n=0}^N0.7^n = \frac{1}{1-0.7}.$$

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, $$s_N = \sum_{n=0}^N 0.7^n.$$ 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 $$\begin{array}{rcrcrcrcrcrcl} s_N &= 1 &+& (0.7) &+& (0.7)^2 &+& \cdots &+& (0.7)^N\\ (0.7)s_N &= &&(0.7) &+& (0.7)^2 &+&\cdots &+&(0.7)^N &+& (0.7)^{N+1} \end{array}$$ so that $$(1-0.7)s_N = s_N - (0.7)s_N = 1 - (0.7)^{N+1}.$$ Solving for $s_N$ gives $$s_N = \frac{1 - (0.7)^{N+1}}{1-0.7}.$$ 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, $$\lim_{N\to\infty}s_N = \lim_{N\to\infty}\left(\frac{1-(0.7)^{N+1}}{1-0.7}\right) = \frac{\lim\limits_{N\to\infty}1 - \lim\limits_{N\to\infty}(0.7)^{N+1}}{\lim\limits_{N\to\infty}1 - \lim\limits_{N\to\infty}0.7} = \frac{1 - 0}{1-0.7} = \frac{1}{1-0.7}.$$ Since the limit exists, then we write $$\sum_{n=0}^{\infty}(0.7)^n = \frac{1}{1-0.7}.$$

More generally, a sum of the form $$a + ar + ar^2 + ar^3 + \cdots + ar^k$$ 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 $$S = a +ar + \cdots + ar^k$$ we have that $$S - rS = (a+ar+\cdots+ar^k) - (ar+ar^2+\cdots+a^{k+1}) = a - ar^{k+1}$$ so that $$(1-r)S = a(1 - r^{k+1}).$$ 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 $$ \sum_{n=0}^{\infty}ar^{n} $$ with $a$ and $r$ constants is called an infinite geometric series. If $r=1$, then $$ \lim_{N\to\infty}\sum_{n=0}^{N}ar^{n} = \lim_{N\to\infty}\sum_{n=0}^{N}a = \lim_{N\to\infty}(N+1)a = \infty, $$ 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 $$ \lim_{N\to\infty}r^{N+1} = \left\{\begin{array}{ll} 0 &\mbox{if $|r|\lt 1$;}\\ 1 & \mbox{if $r=1$;}\\ \text{does not exist} &\mbox{if $r=-1$ or $|r|\gt 1$} \end{array}\right. $$ it follows that: $$ \begin{align*} \sum_{n=0}^{\infty}ar^{n} &=\left\{\begin{array}{ll} 0 &\mbox{if $a=0$;}\\ \text{diverges}&\mbox{if $a\neq 0$ and $r=1$;}\\ \lim\limits_{N\to\infty}\frac{a(1-r^{N+1})}{1-r} &\mbox{if $r\neq 1$;}\end{array}\right.\\ &= \left\{\begin{array}{ll} \text{diverges}&\mbox{if $a\neq 0$ and $r=1$;}\\ \text{diverges}&\mbox{if $a\neq 0$, and $r=-1$ or $|r|\gt 1$;}\\ \frac{a(1-0)}{1-r}&\mbox{if $|r|\lt 1$;} \end{array}\right.\\ &=\left\{\begin{array}{ll} \text{diverges}&\mbox{if $a\neq 0$ and $|r|\geq 1$;}\\ \frac{a}{1-r}&\mbox{if $|r|\lt 1$.} \end{array}\right. \end{align*} $$

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

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


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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    $\begingroup$ I think we should use this question for a generalization. The above answer already has an answer to the general version. $\endgroup$ – Aryabhata Mar 31 '11 at 16:58
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    $\begingroup$ Arturo, I will mention one more way of proving that $r^n$ converges to $0$ as $n \to \infty$ if $|r| < 1$ that looks somewhat similar to your proof in the divergent case. Take $|r| = \frac{1}{1+a}$ for some $a > 0$. Then $$|r^n - 0| = |r|^n = \frac{1}{(1+a)^n} \leq \frac{1}{1+an} \leq \frac{1}{an} ,$$ which can be made sufficiently small by taking $n$ large enough. $\endgroup$ – Srivatsan Sep 30 '11 at 4:13
  • $\begingroup$ When you say that the limit of the partial sums of a geometric series exists iff the limit of $r^{N+1}$ exists, that's wrong, because it still diverges (as you state later) when $r=1$. $\endgroup$ – Loki Clock Jun 7 '13 at 21:49
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it's called a geometric series. let $-1<x<1$ and let $S_n=1+x+x^2+...+x^n$. then $xS_n=x+...+x^{n+1}=S^n-1+x^{n+1}$. move stuff around to get $$S_n=\frac{1-x^{n+1}}{1-x}$$ take the limit as $n\to\infty$ (noting that $x^n\to0$ if $|x|<1$)

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  • $\begingroup$ In yoyo's answer, you should also take into account that $x^{n+1}\to 0$ as $n\to\infty$ if $|x| < 1$. $\endgroup$ – knightofmathematics Mar 25 '11 at 16:48
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By binomial theorem,

$$\frac{1}{1-x}=(1-x)^{-1}\\=1+{-1\choose1}(-x)+{-1\choose2}(-x)^2+{-1\choose3}(-x)^3...\\=1+x+x^2+x^3+...,$$

which is a geometric series.

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    $\begingroup$ Haha, a unique little jewel in the rough that ought've had more attention. $\endgroup$ – Simply Beautiful Art Dec 30 '16 at 16:35
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If u expand your summation you get series $$1+0.7+0.7^2+\dots$$ as it is geometric series (you can see it here http://en.wikipedia.org/wiki/Geometric_series) $$\sum_{n=0}^{\infty}{0.7^n}=\frac{1}{1-0.7}$$

Or $S=1+x+x^2+\dots$

$xS_n=x+x^2+\dots=S-1$ now u take

$S-xS=1$

$S(1-x)=1 \implies S=\frac{1}{1-x}$

here your $x=0.7$

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I find the proof here lovely.

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  • $\begingroup$ @BabyDragon Here is the updated link. $\endgroup$ – David Mitra Oct 6 '13 at 9:12
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Using the fact that $$x^n - y^n=(x-y)\sum_{k=1}^nx^{k-1}y^{n-k}$$

Set $x=1$ and $y \in \mathbb{N}$ to get the desired result.

$$1 - y^n=(1-y)\sum_{k=1}^ny^{n-k} \iff \sum_{k=1}^ny^{n-k}=y^n\sum_{k=1}^ny^{-k}=\frac{y^n-1}{y-1} \iff \sum_{k=1}^n \frac{1}{y^k} = \frac{y^n-1}{y^{n+1}-y^n}$$

Now, using the limit of the result we have $$\large{\lim_{n\to \infty}\frac{y^n-1}{y^{n+1}-y^n} = \frac{1}{y-1}}$$.

Some examples,

$$\sum_{k=1}^\infty \left(\frac{1}{2}\right)^k = 1$$ $$\sum_{k=1}^\infty \left(\frac{1}{3}\right)^k = 1/2$$ $$\sum_{k=1}^\infty \left(\frac{1}{4}\right)^k = 1/3$$ $$\sum_{k=1}^\infty \left(\frac{1}{5}\right)^k = 1/4$$

Also, if $y \to \infty \implies $

$$\sum_{k=1}^\infty \left(\frac{1}{y}\right)^k = 0$$

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