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Proving convergence of geometric series by induction
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Let $$S = sumlimits_0^infty x^j,$$
where $|x|<1$. I want to show by induction that $$S=frac11-x.$$
But I'm stuck in my attempts to come up with an induction definition.
For example, $S_j = frac11-x-x^j$ clearly does not work here. Should I seek a recursive definition instead?
real-analysis sequences-and-series induction
asked Jul 19 at 3:20
sequence
4,03711031
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up vote
1
down vote
favorite
Let $$S = sumlimits_0^infty x^j,$$
where $|x|<1$. I want to show by induction that $$S=frac11-x.$$
But I'm stuck in my attempts to come up with an induction definition.
For example, $S_j = frac11-x-x^j$ clearly does not work here. Should I seek a recursive definition instead?
real-analysis sequences-and-series induction
asked Jul 19 at 3:20
sequence
4,03711031
6
Try to prove (by induction for instance) that the partial sum $sum_j=0^Nx^j$ equals $frac1-x^N+11-x$ whenever $x$ is different from $1$. Once you have this, take the limit.
– Suzet
Jul 19 at 3:22
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $$S = sumlimits_0^infty x^j,$$
where $|x|<1$. I want to show by induction that $$S=frac11-x.$$
But I'm stuck in my attempts to come up with an induction definition.
For example, $S_j = frac11-x-x^j$ clearly does not work here. Should I seek a recursive definition instead?
real-analysis sequences-and-series induction
asked Jul 19 at 3:20
sequence
4,03711031
Let $$S = sumlimits_0^infty x^j,$$
where $|x|<1$. I want to show by induction that $$S=frac11-x.$$
But I'm stuck in my attempts to come up with an induction definition.
For example, $S_j = frac11-x-x^j$ clearly does not work here. Should I seek a recursive definition instead?
real-analysis sequences-and-series induction
asked Jul 19 at 3:20
sequence
4,03711031
asked Jul 19 at 3:20
sequence
4,03711031
asked Jul 19 at 3:20
sequence
4,03711031
asked Jul 19 at 3:20
sequence
4,03711031
4,03711031
6
Try to prove (by induction for instance) that the partial sum $sum_j=0^Nx^j$ equals $frac1-x^N+11-x$ whenever $x$ is different from $1$. Once you have this, take the limit.
– Suzet
Jul 19 at 3:22
add a comment |Â
6
Try to prove (by induction for instance) that the partial sum $sum_j=0^Nx^j$ equals $frac1-x^N+11-x$ whenever $x$ is different from $1$. Once you have this, take the limit.
– Suzet
Jul 19 at 3:22
6
6
Try to prove (by induction for instance) that the partial sum $sum_j=0^Nx^j$ equals $frac1-x^N+11-x$ whenever $x$ is different from $1$. Once you have this, take the limit.
– Suzet
Jul 19 at 3:22
Try to prove (by induction for instance) that the partial sum $sum_j=0^Nx^j$ equals $frac1-x^N+11-x$ whenever $x$ is different from $1$. Once you have this, take the limit.
– Suzet
Jul 19 at 3:22
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
You should show that for every finite $N$ from induction that beginalign sum_k=0^N r^k = frac1-r^N+11-r endalign then it follows that this limit converges if and only if $r^N+1$ converges to a finite non-1(so that the denominator is well defined) value, which happens if and only if $|r| < 1$.
But here's a much more short proof: beginalignS_N=sum_k=0^N r^k = 1 + r + r^2 + ... + r^N endalign
beginalign (1-r)S_N = 1 + r + r^2 +... + r^N - r - r^2 - ... -r^N+1 = 1 - r^N+1 endalign This implies that $S_N = frac1-r^N+11-r$. Then let $N rightarrow infty$ to see it converges if and only if $|r| < 1$
answered Jul 19 at 4:09
Raymond Chu
1,03719
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You should show that for every finite $N$ from induction that beginalign sum_k=0^N r^k = frac1-r^N+11-r endalign then it follows that this limit converges if and only if $r^N+1$ converges to a finite non-1(so that the denominator is well defined) value, which happens if and only if $|r| < 1$.
But here's a much more short proof: beginalignS_N=sum_k=0^N r^k = 1 + r + r^2 + ... + r^N endalign
beginalign (1-r)S_N = 1 + r + r^2 +... + r^N - r - r^2 - ... -r^N+1 = 1 - r^N+1 endalign This implies that $S_N = frac1-r^N+11-r$. Then let $N rightarrow infty$ to see it converges if and only if $|r| < 1$
answered Jul 19 at 4:09
Raymond Chu
1,03719
add a comment |Â
up vote
3
down vote
accepted
You should show that for every finite $N$ from induction that beginalign sum_k=0^N r^k = frac1-r^N+11-r endalign then it follows that this limit converges if and only if $r^N+1$ converges to a finite non-1(so that the denominator is well defined) value, which happens if and only if $|r| < 1$.
But here's a much more short proof: beginalignS_N=sum_k=0^N r^k = 1 + r + r^2 + ... + r^N endalign
beginalign (1-r)S_N = 1 + r + r^2 +... + r^N - r - r^2 - ... -r^N+1 = 1 - r^N+1 endalign This implies that $S_N = frac1-r^N+11-r$. Then let $N rightarrow infty$ to see it converges if and only if $|r| < 1$
answered Jul 19 at 4:09
Raymond Chu
1,03719
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You should show that for every finite $N$ from induction that beginalign sum_k=0^N r^k = frac1-r^N+11-r endalign then it follows that this limit converges if and only if $r^N+1$ converges to a finite non-1(so that the denominator is well defined) value, which happens if and only if $|r| < 1$.
But here's a much more short proof: beginalignS_N=sum_k=0^N r^k = 1 + r + r^2 + ... + r^N endalign
beginalign (1-r)S_N = 1 + r + r^2 +... + r^N - r - r^2 - ... -r^N+1 = 1 - r^N+1 endalign This implies that $S_N = frac1-r^N+11-r$. Then let $N rightarrow infty$ to see it converges if and only if $|r| < 1$
answered Jul 19 at 4:09
Raymond Chu
1,03719
You should show that for every finite $N$ from induction that beginalign sum_k=0^N r^k = frac1-r^N+11-r endalign then it follows that this limit converges if and only if $r^N+1$ converges to a finite non-1(so that the denominator is well defined) value, which happens if and only if $|r| < 1$.
But here's a much more short proof: beginalignS_N=sum_k=0^N r^k = 1 + r + r^2 + ... + r^N endalign
beginalign (1-r)S_N = 1 + r + r^2 +... + r^N - r - r^2 - ... -r^N+1 = 1 - r^N+1 endalign This implies that $S_N = frac1-r^N+11-r$. Then let $N rightarrow infty$ to see it converges if and only if $|r| < 1$
answered Jul 19 at 4:09
Raymond Chu
1,03719
answered Jul 19 at 4:09
Raymond Chu
1,03719
answered Jul 19 at 4:09
Raymond Chu
1,03719
answered Jul 19 at 4:09
Raymond Chu
1,03719
1,03719
add a comment |Â
add a comment |Â
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6
Try to prove (by induction for instance) that the partial sum $sum_j=0^Nx^j$ equals $frac1-x^N+11-x$ whenever $x$ is different from $1$. Once you have this, take the limit.
– Suzet
Jul 19 at 3:22