Mixing
Finding limit of a strangely defined recursive sequence
I have problems finding limits of recursive sequence like: $a_1 = 5$ $a_{n+1} = frac{(a_n)^2}{10n + 3} + 1$
I know that intuitively the limit will be $1$, since $frac{(a_n)^2}{10n + 3}$ tends to 0, but I have problems proving that.
Hints would be appreciated.
limits limits-without-lhopital
asked Nov 22 '18 at 10:29
JoséJosé
11
add a comment |
I have problems finding limits of recursive sequence like: $a_1 = 5$ $a_{n+1} = frac{(a_n)^2}{10n + 3} + 1$
I know that intuitively the limit will be $1$, since $frac{(a_n)^2}{10n + 3}$ tends to 0, but I have problems proving that.
Hints would be appreciated.
limits limits-without-lhopital
asked Nov 22 '18 at 10:29
JoséJosé
11
add a comment |
I have problems finding limits of recursive sequence like: $a_1 = 5$ $a_{n+1} = frac{(a_n)^2}{10n + 3} + 1$
I know that intuitively the limit will be $1$, since $frac{(a_n)^2}{10n + 3}$ tends to 0, but I have problems proving that.
Hints would be appreciated.
limits limits-without-lhopital
asked Nov 22 '18 at 10:29
JoséJosé
11
I have problems finding limits of recursive sequence like: $a_1 = 5$ $a_{n+1} = frac{(a_n)^2}{10n + 3} + 1$
I know that intuitively the limit will be $1$, since $frac{(a_n)^2}{10n + 3}$ tends to 0, but I have problems proving that.
Hints would be appreciated.
limits limits-without-lhopital
limits limits-without-lhopital
asked Nov 22 '18 at 10:29
JoséJosé
11
asked Nov 22 '18 at 10:29
JoséJosé
11
asked Nov 22 '18 at 10:29
JoséJosé
11
asked Nov 22 '18 at 10:29
JoséJosé
11
asked Nov 22 '18 at 10:29
JoséJosé
11
11
add a comment |
add a comment |
2 Answers
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First to all we'll need to show that $a_n$ is monotone, in particular decreasing by induction.
It's simple verify that $a_2<a_1$ and $a_ngeq 1$ for every $n$ because the term $frac{a^2_n}{10n+3}$ is always $geq 0$, now suppose $a_n<a_{n-1}$ then $a_n^2<a_{n-1}^2$ and
$$
a_{n+1}=frac{a_n^2}{10(n+1)+3}+1<frac{a_{n-1}^2}{10(n+1)+3}+1<frac{a_{n-1}^2}{10n+3}+1=a_n
$$
Now $a_n$ is decreasing and lower bounded by $1$ so $a_n$ has got a limit: $lim_{nrightarrow +infty}a_n=linmathbb R$. Then
$$
l=lim_{nrightarrow +infty}a_{n+1}=lim_{nrightarrow +infty}frac{a_n^2}{10n+3}+1=frac l{+infty}+1=0+1=1
$$
then $l=1$.
answered Nov 22 '18 at 10:50
P De DonatoP De Donato
4147
Would it be enough to show that $a_n$ is non-increasing instead of showing it's decreasing?
– José
Nov 22 '18 at 11:14
Yes, for generic successions $a_n$ you need at least one of: 1. $a_n$ not decreasing ($a_ngeq a_{n-1}$) and upper bounded; 2. $a_n$ not increasing ($a_nleq a_{n-1}$) and lower bounded.
– P De Donato
Nov 22 '18 at 11:16
For example succession $a_n=(-1)^n$ is lower and upper bounded but hasn't limit (it isn't monotone).
– P De Donato
Nov 22 '18 at 11:19
add a comment |
It should be clear that $a_n >0$ for all n.
Now prove, by induction, that $(a_n)$ is decreasing.
Hence $(a_n)$ is monotonic and bounded and therefore convergent. If $a$ is the limit of $(a_n)$ then $ frac{(a_n)^2}{10n + 3} + 1 to 1$ as $n to infty$, hence $a_{n+1} to 1$ as $n to infty$.
answered Nov 22 '18 at 10:48
FredFred
44.4k1845
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
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First to all we'll need to show that $a_n$ is monotone, in particular decreasing by induction.
It's simple verify that $a_2<a_1$ and $a_ngeq 1$ for every $n$ because the term $frac{a^2_n}{10n+3}$ is always $geq 0$, now suppose $a_n<a_{n-1}$ then $a_n^2<a_{n-1}^2$ and
$$
a_{n+1}=frac{a_n^2}{10(n+1)+3}+1<frac{a_{n-1}^2}{10(n+1)+3}+1<frac{a_{n-1}^2}{10n+3}+1=a_n
$$
Now $a_n$ is decreasing and lower bounded by $1$ so $a_n$ has got a limit: $lim_{nrightarrow +infty}a_n=linmathbb R$. Then
$$
l=lim_{nrightarrow +infty}a_{n+1}=lim_{nrightarrow +infty}frac{a_n^2}{10n+3}+1=frac l{+infty}+1=0+1=1
$$
then $l=1$.
answered Nov 22 '18 at 10:50
P De DonatoP De Donato
4147
Would it be enough to show that $a_n$ is non-increasing instead of showing it's decreasing?
– José
Nov 22 '18 at 11:14
Yes, for generic successions $a_n$ you need at least one of: 1. $a_n$ not decreasing ($a_ngeq a_{n-1}$) and upper bounded; 2. $a_n$ not increasing ($a_nleq a_{n-1}$) and lower bounded.
– P De Donato
Nov 22 '18 at 11:16
For example succession $a_n=(-1)^n$ is lower and upper bounded but hasn't limit (it isn't monotone).
– P De Donato
Nov 22 '18 at 11:19
add a comment |
First to all we'll need to show that $a_n$ is monotone, in particular decreasing by induction.
It's simple verify that $a_2<a_1$ and $a_ngeq 1$ for every $n$ because the term $frac{a^2_n}{10n+3}$ is always $geq 0$, now suppose $a_n<a_{n-1}$ then $a_n^2<a_{n-1}^2$ and
$$
a_{n+1}=frac{a_n^2}{10(n+1)+3}+1<frac{a_{n-1}^2}{10(n+1)+3}+1<frac{a_{n-1}^2}{10n+3}+1=a_n
$$
Now $a_n$ is decreasing and lower bounded by $1$ so $a_n$ has got a limit: $lim_{nrightarrow +infty}a_n=linmathbb R$. Then
$$
l=lim_{nrightarrow +infty}a_{n+1}=lim_{nrightarrow +infty}frac{a_n^2}{10n+3}+1=frac l{+infty}+1=0+1=1
$$
then $l=1$.
answered Nov 22 '18 at 10:50
P De DonatoP De Donato
4147
Would it be enough to show that $a_n$ is non-increasing instead of showing it's decreasing?
– José
Nov 22 '18 at 11:14
Yes, for generic successions $a_n$ you need at least one of: 1. $a_n$ not decreasing ($a_ngeq a_{n-1}$) and upper bounded; 2. $a_n$ not increasing ($a_nleq a_{n-1}$) and lower bounded.
– P De Donato
Nov 22 '18 at 11:16
For example succession $a_n=(-1)^n$ is lower and upper bounded but hasn't limit (it isn't monotone).
– P De Donato
Nov 22 '18 at 11:19
add a comment |
First to all we'll need to show that $a_n$ is monotone, in particular decreasing by induction.
It's simple verify that $a_2<a_1$ and $a_ngeq 1$ for every $n$ because the term $frac{a^2_n}{10n+3}$ is always $geq 0$, now suppose $a_n<a_{n-1}$ then $a_n^2<a_{n-1}^2$ and
$$
a_{n+1}=frac{a_n^2}{10(n+1)+3}+1<frac{a_{n-1}^2}{10(n+1)+3}+1<frac{a_{n-1}^2}{10n+3}+1=a_n
$$
Now $a_n$ is decreasing and lower bounded by $1$ so $a_n$ has got a limit: $lim_{nrightarrow +infty}a_n=linmathbb R$. Then
$$
l=lim_{nrightarrow +infty}a_{n+1}=lim_{nrightarrow +infty}frac{a_n^2}{10n+3}+1=frac l{+infty}+1=0+1=1
$$
then $l=1$.
answered Nov 22 '18 at 10:50
P De DonatoP De Donato
4147
First to all we'll need to show that $a_n$ is monotone, in particular decreasing by induction.
It's simple verify that $a_2<a_1$ and $a_ngeq 1$ for every $n$ because the term $frac{a^2_n}{10n+3}$ is always $geq 0$, now suppose $a_n<a_{n-1}$ then $a_n^2<a_{n-1}^2$ and
$$
a_{n+1}=frac{a_n^2}{10(n+1)+3}+1<frac{a_{n-1}^2}{10(n+1)+3}+1<frac{a_{n-1}^2}{10n+3}+1=a_n
$$
Now $a_n$ is decreasing and lower bounded by $1$ so $a_n$ has got a limit: $lim_{nrightarrow +infty}a_n=linmathbb R$. Then
$$
l=lim_{nrightarrow +infty}a_{n+1}=lim_{nrightarrow +infty}frac{a_n^2}{10n+3}+1=frac l{+infty}+1=0+1=1
$$
then $l=1$.
answered Nov 22 '18 at 10:50
P De DonatoP De Donato
4147
answered Nov 22 '18 at 10:50
P De DonatoP De Donato
4147
answered Nov 22 '18 at 10:50
P De DonatoP De Donato
4147
answered Nov 22 '18 at 10:50
P De DonatoP De Donato
4147
4147
Would it be enough to show that $a_n$ is non-increasing instead of showing it's decreasing?
– José
Nov 22 '18 at 11:14
Yes, for generic successions $a_n$ you need at least one of: 1. $a_n$ not decreasing ($a_ngeq a_{n-1}$) and upper bounded; 2. $a_n$ not increasing ($a_nleq a_{n-1}$) and lower bounded.
– P De Donato
Nov 22 '18 at 11:16
For example succession $a_n=(-1)^n$ is lower and upper bounded but hasn't limit (it isn't monotone).
– P De Donato
Nov 22 '18 at 11:19
add a comment |
Would it be enough to show that $a_n$ is non-increasing instead of showing it's decreasing?
– José
Nov 22 '18 at 11:14
Yes, for generic successions $a_n$ you need at least one of: 1. $a_n$ not decreasing ($a_ngeq a_{n-1}$) and upper bounded; 2. $a_n$ not increasing ($a_nleq a_{n-1}$) and lower bounded.
– P De Donato
Nov 22 '18 at 11:16
For example succession $a_n=(-1)^n$ is lower and upper bounded but hasn't limit (it isn't monotone).
– P De Donato
Nov 22 '18 at 11:19
Would it be enough to show that $a_n$ is non-increasing instead of showing it's decreasing?
– José
Nov 22 '18 at 11:14
Would it be enough to show that $a_n$ is non-increasing instead of showing it's decreasing?
– José
Nov 22 '18 at 11:14
Yes, for generic successions $a_n$ you need at least one of: 1. $a_n$ not decreasing ($a_ngeq a_{n-1}$) and upper bounded; 2. $a_n$ not increasing ($a_nleq a_{n-1}$) and lower bounded.
– P De Donato
Nov 22 '18 at 11:16
Yes, for generic successions $a_n$ you need at least one of: 1. $a_n$ not decreasing ($a_ngeq a_{n-1}$) and upper bounded; 2. $a_n$ not increasing ($a_nleq a_{n-1}$) and lower bounded.
– P De Donato
Nov 22 '18 at 11:16
For example succession $a_n=(-1)^n$ is lower and upper bounded but hasn't limit (it isn't monotone).
– P De Donato
Nov 22 '18 at 11:19
For example succession $a_n=(-1)^n$ is lower and upper bounded but hasn't limit (it isn't monotone).
– P De Donato
Nov 22 '18 at 11:19
add a comment |
It should be clear that $a_n >0$ for all n.
Now prove, by induction, that $(a_n)$ is decreasing.
Hence $(a_n)$ is monotonic and bounded and therefore convergent. If $a$ is the limit of $(a_n)$ then $ frac{(a_n)^2}{10n + 3} + 1 to 1$ as $n to infty$, hence $a_{n+1} to 1$ as $n to infty$.
answered Nov 22 '18 at 10:48
FredFred
44.4k1845
add a comment |
It should be clear that $a_n >0$ for all n.
Now prove, by induction, that $(a_n)$ is decreasing.
Hence $(a_n)$ is monotonic and bounded and therefore convergent. If $a$ is the limit of $(a_n)$ then $ frac{(a_n)^2}{10n + 3} + 1 to 1$ as $n to infty$, hence $a_{n+1} to 1$ as $n to infty$.
answered Nov 22 '18 at 10:48
FredFred
44.4k1845
add a comment |
It should be clear that $a_n >0$ for all n.
Now prove, by induction, that $(a_n)$ is decreasing.
Hence $(a_n)$ is monotonic and bounded and therefore convergent. If $a$ is the limit of $(a_n)$ then $ frac{(a_n)^2}{10n + 3} + 1 to 1$ as $n to infty$, hence $a_{n+1} to 1$ as $n to infty$.
answered Nov 22 '18 at 10:48
FredFred
44.4k1845
It should be clear that $a_n >0$ for all n.
Now prove, by induction, that $(a_n)$ is decreasing.
Hence $(a_n)$ is monotonic and bounded and therefore convergent. If $a$ is the limit of $(a_n)$ then $ frac{(a_n)^2}{10n + 3} + 1 to 1$ as $n to infty$, hence $a_{n+1} to 1$ as $n to infty$.
answered Nov 22 '18 at 10:48
FredFred
44.4k1845
answered Nov 22 '18 at 10:48
FredFred
44.4k1845
answered Nov 22 '18 at 10:48
FredFred
44.4k1845
answered Nov 22 '18 at 10:48
FredFred
44.4k1845
44.4k1845
add a comment |
add a comment |
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