Summation equivalence
$begingroup$
I am having trouble seeing how this summation equivalence holds true:
$sum_{i=0}^infty x^i$ = $frac{1}{1-x}$ if |x| < 1
The only thing I can see where there would be a problem is if x = 1 or x = -1
Im hoping someone can provide me with a clear explanation as I am very confused
proof-verification summation induction
asked Dec 5 '18 at 2:11
ResalexResalex
61
$endgroup$
add a comment |
$begingroup$
I am having trouble seeing how this summation equivalence holds true:
$sum_{i=0}^infty x^i$ = $frac{1}{1-x}$ if |x| < 1
The only thing I can see where there would be a problem is if x = 1 or x = -1
Im hoping someone can provide me with a clear explanation as I am very confused
proof-verification summation induction
asked Dec 5 '18 at 2:11
ResalexResalex
61
$endgroup$
add a comment |
$begingroup$
I am having trouble seeing how this summation equivalence holds true:
$sum_{i=0}^infty x^i$ = $frac{1}{1-x}$ if |x| < 1
The only thing I can see where there would be a problem is if x = 1 or x = -1
Im hoping someone can provide me with a clear explanation as I am very confused
proof-verification summation induction
asked Dec 5 '18 at 2:11
ResalexResalex
61
$endgroup$
I am having trouble seeing how this summation equivalence holds true:
$sum_{i=0}^infty x^i$ = $frac{1}{1-x}$ if |x| < 1
The only thing I can see where there would be a problem is if x = 1 or x = -1
Im hoping someone can provide me with a clear explanation as I am very confused
proof-verification summation induction
proof-verification summation induction
asked Dec 5 '18 at 2:11
ResalexResalex
61
asked Dec 5 '18 at 2:11
ResalexResalex
61
asked Dec 5 '18 at 2:11
ResalexResalex
61
asked Dec 5 '18 at 2:11
ResalexResalex
61
asked Dec 5 '18 at 2:11
ResalexResalex
61
61
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I actually thought about this earlier and came up with a somewhat heuristic argument. So we know, for a geometric series with ratio $r$,
$$sum_{n=0}^k r^n = 1 + r + r^2 + r^3 + ... + r^k = frac{r^{k+1}-1}{r-1} = frac{1 - r^{k+1}}{1-r}$$
(Either of the fractions on the right are valid; which you choose is merely a matter of personal taste. You can see their equivalence by multiplying the top and bottom by $-1$. We will be considering the latter however since it's closer to the formula that most people see.)
Consider the limiting behavior of this as $k to infty$, i.e. as we accumulate infinitely many terms. What happens? Well, we have...
$$sum_{n=0}^infty r^n = lim_{k to infty} frac{1 - r^{k+1}}{1-r}$$
Well, suppose $r = 1$. Well that's dumb, the limit's undefined, so it's really hard to observe anything there. (Of course, we can show that the limit would be infinity by instead considering the limit of the partial sums.)
Suppose $|r| > 1$. Then clearly, owing to the $r^{k+1}$ term, the fractions increases in magnitude without bound, i.e. diverges. So not much there.
But what if $|r| < 1$? Then $r^{k+1} to 0$ - it shouldn't be difficult to convince yourself of that much.
Thus, we say, for $|r|<1$,
$$sum_{n=0}^infty r^n = frac{1}{1-r}$$
and for $|r| geq 1$, it diverges.
answered Dec 5 '18 at 2:24
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
add a comment |
$begingroup$
Consider the formula for the sum of a finite geometric series:
$$ sum_{i=0}^n x^i = frac{1-x^n}{1-x}.$$
Now, if we have that the $|x| < 1$, then when we repeatedly multiply $x$ by itself, it gets smaller. This may give some intuition as to why we can then let $n$ tend to infinity in this formula to give
$$ sum_{i=1}^infty x^i = frac{1}{1-x},$$ but only when $|x| < 1$, as then $x^n$ tends to 0 as $n$ tends to infinity. We say that 1 is the radius of convergence of the sum - note that $1$ is not included in this radius, so the sum is undefined when $x$ is 1. This should make sense, since $1 + 1 + 1 + cdots $ intuitively doesn't seem to look like it might end up being finite, right?
answered Dec 5 '18 at 2:21
Stuartg98Stuartg98
586
$endgroup$
$begingroup$
isnt it undefined because when x is 1 you are dividng by 0?
$endgroup$
– Resalex
Dec 5 '18 at 2:23
$begingroup$
Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
$endgroup$
– Stuartg98
Dec 5 '18 at 2:25
$begingroup$
Does that make sense?
$endgroup$
– Stuartg98
Dec 5 '18 at 2:26
1
$begingroup$
ah okay yes it does thank you very much
$endgroup$
– Resalex
Dec 5 '18 at 2:26
$begingroup$
No problem! Glad I could help.
$endgroup$
– Stuartg98
Dec 5 '18 at 2:28
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
I actually thought about this earlier and came up with a somewhat heuristic argument. So we know, for a geometric series with ratio $r$,
$$sum_{n=0}^k r^n = 1 + r + r^2 + r^3 + ... + r^k = frac{r^{k+1}-1}{r-1} = frac{1 - r^{k+1}}{1-r}$$
(Either of the fractions on the right are valid; which you choose is merely a matter of personal taste. You can see their equivalence by multiplying the top and bottom by $-1$. We will be considering the latter however since it's closer to the formula that most people see.)
Consider the limiting behavior of this as $k to infty$, i.e. as we accumulate infinitely many terms. What happens? Well, we have...
$$sum_{n=0}^infty r^n = lim_{k to infty} frac{1 - r^{k+1}}{1-r}$$
Well, suppose $r = 1$. Well that's dumb, the limit's undefined, so it's really hard to observe anything there. (Of course, we can show that the limit would be infinity by instead considering the limit of the partial sums.)
Suppose $|r| > 1$. Then clearly, owing to the $r^{k+1}$ term, the fractions increases in magnitude without bound, i.e. diverges. So not much there.
But what if $|r| < 1$? Then $r^{k+1} to 0$ - it shouldn't be difficult to convince yourself of that much.
Thus, we say, for $|r|<1$,
$$sum_{n=0}^infty r^n = frac{1}{1-r}$$
and for $|r| geq 1$, it diverges.
answered Dec 5 '18 at 2:24
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
add a comment |
$begingroup$
I actually thought about this earlier and came up with a somewhat heuristic argument. So we know, for a geometric series with ratio $r$,
$$sum_{n=0}^k r^n = 1 + r + r^2 + r^3 + ... + r^k = frac{r^{k+1}-1}{r-1} = frac{1 - r^{k+1}}{1-r}$$
(Either of the fractions on the right are valid; which you choose is merely a matter of personal taste. You can see their equivalence by multiplying the top and bottom by $-1$. We will be considering the latter however since it's closer to the formula that most people see.)
Consider the limiting behavior of this as $k to infty$, i.e. as we accumulate infinitely many terms. What happens? Well, we have...
$$sum_{n=0}^infty r^n = lim_{k to infty} frac{1 - r^{k+1}}{1-r}$$
Well, suppose $r = 1$. Well that's dumb, the limit's undefined, so it's really hard to observe anything there. (Of course, we can show that the limit would be infinity by instead considering the limit of the partial sums.)
Suppose $|r| > 1$. Then clearly, owing to the $r^{k+1}$ term, the fractions increases in magnitude without bound, i.e. diverges. So not much there.
But what if $|r| < 1$? Then $r^{k+1} to 0$ - it shouldn't be difficult to convince yourself of that much.
Thus, we say, for $|r|<1$,
$$sum_{n=0}^infty r^n = frac{1}{1-r}$$
and for $|r| geq 1$, it diverges.
answered Dec 5 '18 at 2:24
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
add a comment |
$begingroup$
I actually thought about this earlier and came up with a somewhat heuristic argument. So we know, for a geometric series with ratio $r$,
$$sum_{n=0}^k r^n = 1 + r + r^2 + r^3 + ... + r^k = frac{r^{k+1}-1}{r-1} = frac{1 - r^{k+1}}{1-r}$$
(Either of the fractions on the right are valid; which you choose is merely a matter of personal taste. You can see their equivalence by multiplying the top and bottom by $-1$. We will be considering the latter however since it's closer to the formula that most people see.)
Consider the limiting behavior of this as $k to infty$, i.e. as we accumulate infinitely many terms. What happens? Well, we have...
$$sum_{n=0}^infty r^n = lim_{k to infty} frac{1 - r^{k+1}}{1-r}$$
Well, suppose $r = 1$. Well that's dumb, the limit's undefined, so it's really hard to observe anything there. (Of course, we can show that the limit would be infinity by instead considering the limit of the partial sums.)
Suppose $|r| > 1$. Then clearly, owing to the $r^{k+1}$ term, the fractions increases in magnitude without bound, i.e. diverges. So not much there.
But what if $|r| < 1$? Then $r^{k+1} to 0$ - it shouldn't be difficult to convince yourself of that much.
Thus, we say, for $|r|<1$,
$$sum_{n=0}^infty r^n = frac{1}{1-r}$$
and for $|r| geq 1$, it diverges.
answered Dec 5 '18 at 2:24
Eevee TrainerEevee Trainer
5,7871936
$endgroup$
I actually thought about this earlier and came up with a somewhat heuristic argument. So we know, for a geometric series with ratio $r$,
$$sum_{n=0}^k r^n = 1 + r + r^2 + r^3 + ... + r^k = frac{r^{k+1}-1}{r-1} = frac{1 - r^{k+1}}{1-r}$$
(Either of the fractions on the right are valid; which you choose is merely a matter of personal taste. You can see their equivalence by multiplying the top and bottom by $-1$. We will be considering the latter however since it's closer to the formula that most people see.)
Consider the limiting behavior of this as $k to infty$, i.e. as we accumulate infinitely many terms. What happens? Well, we have...
$$sum_{n=0}^infty r^n = lim_{k to infty} frac{1 - r^{k+1}}{1-r}$$
Well, suppose $r = 1$. Well that's dumb, the limit's undefined, so it's really hard to observe anything there. (Of course, we can show that the limit would be infinity by instead considering the limit of the partial sums.)
Suppose $|r| > 1$. Then clearly, owing to the $r^{k+1}$ term, the fractions increases in magnitude without bound, i.e. diverges. So not much there.
But what if $|r| < 1$? Then $r^{k+1} to 0$ - it shouldn't be difficult to convince yourself of that much.
Thus, we say, for $|r|<1$,
$$sum_{n=0}^infty r^n = frac{1}{1-r}$$
and for $|r| geq 1$, it diverges.
answered Dec 5 '18 at 2:24
Eevee TrainerEevee Trainer
5,7871936
answered Dec 5 '18 at 2:24
Eevee TrainerEevee Trainer
5,7871936
answered Dec 5 '18 at 2:24
Eevee TrainerEevee Trainer
5,7871936
answered Dec 5 '18 at 2:24
Eevee TrainerEevee Trainer
5,7871936
5,7871936
add a comment |
add a comment |
$begingroup$
Consider the formula for the sum of a finite geometric series:
$$ sum_{i=0}^n x^i = frac{1-x^n}{1-x}.$$
Now, if we have that the $|x| < 1$, then when we repeatedly multiply $x$ by itself, it gets smaller. This may give some intuition as to why we can then let $n$ tend to infinity in this formula to give
$$ sum_{i=1}^infty x^i = frac{1}{1-x},$$ but only when $|x| < 1$, as then $x^n$ tends to 0 as $n$ tends to infinity. We say that 1 is the radius of convergence of the sum - note that $1$ is not included in this radius, so the sum is undefined when $x$ is 1. This should make sense, since $1 + 1 + 1 + cdots $ intuitively doesn't seem to look like it might end up being finite, right?
answered Dec 5 '18 at 2:21
Stuartg98Stuartg98
586
$endgroup$
$begingroup$
isnt it undefined because when x is 1 you are dividng by 0?
$endgroup$
– Resalex
Dec 5 '18 at 2:23
$begingroup$
Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
$endgroup$
– Stuartg98
Dec 5 '18 at 2:25
$begingroup$
Does that make sense?
$endgroup$
– Stuartg98
Dec 5 '18 at 2:26
1
$begingroup$
ah okay yes it does thank you very much
$endgroup$
– Resalex
Dec 5 '18 at 2:26
$begingroup$
No problem! Glad I could help.
$endgroup$
– Stuartg98
Dec 5 '18 at 2:28
add a comment |
$begingroup$
Consider the formula for the sum of a finite geometric series:
$$ sum_{i=0}^n x^i = frac{1-x^n}{1-x}.$$
Now, if we have that the $|x| < 1$, then when we repeatedly multiply $x$ by itself, it gets smaller. This may give some intuition as to why we can then let $n$ tend to infinity in this formula to give
$$ sum_{i=1}^infty x^i = frac{1}{1-x},$$ but only when $|x| < 1$, as then $x^n$ tends to 0 as $n$ tends to infinity. We say that 1 is the radius of convergence of the sum - note that $1$ is not included in this radius, so the sum is undefined when $x$ is 1. This should make sense, since $1 + 1 + 1 + cdots $ intuitively doesn't seem to look like it might end up being finite, right?
answered Dec 5 '18 at 2:21
Stuartg98Stuartg98
586
$endgroup$
$begingroup$
isnt it undefined because when x is 1 you are dividng by 0?
$endgroup$
– Resalex
Dec 5 '18 at 2:23
$begingroup$
Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
$endgroup$
– Stuartg98
Dec 5 '18 at 2:25
$begingroup$
Does that make sense?
$endgroup$
– Stuartg98
Dec 5 '18 at 2:26
1
$begingroup$
ah okay yes it does thank you very much
$endgroup$
– Resalex
Dec 5 '18 at 2:26
$begingroup$
No problem! Glad I could help.
$endgroup$
– Stuartg98
Dec 5 '18 at 2:28
add a comment |
$begingroup$
Consider the formula for the sum of a finite geometric series:
$$ sum_{i=0}^n x^i = frac{1-x^n}{1-x}.$$
Now, if we have that the $|x| < 1$, then when we repeatedly multiply $x$ by itself, it gets smaller. This may give some intuition as to why we can then let $n$ tend to infinity in this formula to give
$$ sum_{i=1}^infty x^i = frac{1}{1-x},$$ but only when $|x| < 1$, as then $x^n$ tends to 0 as $n$ tends to infinity. We say that 1 is the radius of convergence of the sum - note that $1$ is not included in this radius, so the sum is undefined when $x$ is 1. This should make sense, since $1 + 1 + 1 + cdots $ intuitively doesn't seem to look like it might end up being finite, right?
answered Dec 5 '18 at 2:21
Stuartg98Stuartg98
586
$endgroup$
Consider the formula for the sum of a finite geometric series:
$$ sum_{i=0}^n x^i = frac{1-x^n}{1-x}.$$
Now, if we have that the $|x| < 1$, then when we repeatedly multiply $x$ by itself, it gets smaller. This may give some intuition as to why we can then let $n$ tend to infinity in this formula to give
$$ sum_{i=1}^infty x^i = frac{1}{1-x},$$ but only when $|x| < 1$, as then $x^n$ tends to 0 as $n$ tends to infinity. We say that 1 is the radius of convergence of the sum - note that $1$ is not included in this radius, so the sum is undefined when $x$ is 1. This should make sense, since $1 + 1 + 1 + cdots $ intuitively doesn't seem to look like it might end up being finite, right?
answered Dec 5 '18 at 2:21
Stuartg98Stuartg98
586
answered Dec 5 '18 at 2:21
Stuartg98Stuartg98
586
answered Dec 5 '18 at 2:21
Stuartg98Stuartg98
586
answered Dec 5 '18 at 2:21
Stuartg98Stuartg98
586
586
$begingroup$
isnt it undefined because when x is 1 you are dividng by 0?
$endgroup$
– Resalex
Dec 5 '18 at 2:23
$begingroup$
Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
$endgroup$
– Stuartg98
Dec 5 '18 at 2:25
$begingroup$
Does that make sense?
$endgroup$
– Stuartg98
Dec 5 '18 at 2:26
1
$begingroup$
ah okay yes it does thank you very much
$endgroup$
– Resalex
Dec 5 '18 at 2:26
$begingroup$
No problem! Glad I could help.
$endgroup$
– Stuartg98
Dec 5 '18 at 2:28
add a comment |
$begingroup$
isnt it undefined because when x is 1 you are dividng by 0?
$endgroup$
– Resalex
Dec 5 '18 at 2:23
$begingroup$
Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
$endgroup$
– Stuartg98
Dec 5 '18 at 2:25
$begingroup$
Does that make sense?
$endgroup$
– Stuartg98
Dec 5 '18 at 2:26
1
$begingroup$
ah okay yes it does thank you very much
$endgroup$
– Resalex
Dec 5 '18 at 2:26
$begingroup$
No problem! Glad I could help.
$endgroup$
– Stuartg98
Dec 5 '18 at 2:28
$begingroup$
isnt it undefined because when x is 1 you are dividng by 0?
$endgroup$
– Resalex
Dec 5 '18 at 2:23
$begingroup$
isnt it undefined because when x is 1 you are dividng by 0?
$endgroup$
– Resalex
Dec 5 '18 at 2:23
$begingroup$
Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
$endgroup$
– Stuartg98
Dec 5 '18 at 2:25
$begingroup$
Yes exactly - we don't actually divide by 0 though, since we are only asserting that the right hand side equals the left hand side "when $|x| < 1$"
$endgroup$
– Stuartg98
Dec 5 '18 at 2:25
$begingroup$
Does that make sense?
$endgroup$
– Stuartg98
Dec 5 '18 at 2:26
$begingroup$
Does that make sense?
$endgroup$
– Stuartg98
Dec 5 '18 at 2:26
1
1
$begingroup$
ah okay yes it does thank you very much
$endgroup$
– Resalex
Dec 5 '18 at 2:26
$begingroup$
ah okay yes it does thank you very much
$endgroup$
– Resalex
Dec 5 '18 at 2:26
$begingroup$
No problem! Glad I could help.
$endgroup$
– Stuartg98
Dec 5 '18 at 2:28
$begingroup$
No problem! Glad I could help.
$endgroup$
– Stuartg98
Dec 5 '18 at 2:28
add a comment |
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