Mixing
Is it a rigorous way to show that $x_n = 1 + {1^2over 4} + {2^2over 4^2} + cdots + {n^2over 4^n}$ is...
$begingroup$
Prove the following sequence is convergent:
$$
x_n = 1 + {1^2over 4} + {2^2over 4^2} + cdots + {n^2over 4^n}
$$
I've decided to try Cauchy Criterion here. Consider $|x_m - x_n|$ for $m = 2n > n$:
$$
begin{align}
|x_m - x_n| &= left| 1+ sum_{k=1}^mfrac{k^2}{4^k} - 1 - sum_{k=1}^nfrac{k^2}{4^k}right|\
&= left|sum_{k=n+1}^{2n}frac{k^2}{4^k}right| \
&stackrel{>0}{=} sum_{k=n+1}^{2n}frac{k^2}{4^k} \
&le sum_{k=n+1}^{2n}frac{(2n)^2}{4^k} \
&le sum_{k=n+1}^{2n}frac{(2n)^2}{4^{n+1}} \
&= frac{4n^3}{4^{n+1}} \
&= frac{n^3}{4^n}
end{align}
$$
Consider the limit:
$$
0 le lim_{ntoinfty}|x_{2n} - x_n| le lim_{ntoinfty} frac{n^3}{4^n} = 0
$$
Which by squeeze theorem gives:
$$
lim_{ntoinfty}|x_{2n} - x_n| = 0
$$
proving $x_n$ is a fundamental sequence, hence convergent.
I've been also thinking of using monotone convergence theorem:
$$
x_{n+1} - x_n = frac{(n+1)^2}{4^{n+1}} > 0
$$
By this $x_n$ is monotonically increasing, but i couldn't find an upper bound.
I would like to know two things:
- It the above a rigorous way to show $x_n$ is convergent?
- Even though the problem statement only asks to prove convergence rather than find the limit I would still want to find it. W|A suggests its ${20over 27}$. Is it possible to obtain that result using more or less elementary calculus reasoning?(anything before the definition of a derivative)
calculus sequences-and-series limits proof-verification
asked Jan 9 at 15:44
romanroman
2,17321224
$endgroup$
add a comment |
$begingroup$
Prove the following sequence is convergent:
$$
x_n = 1 + {1^2over 4} + {2^2over 4^2} + cdots + {n^2over 4^n}
$$
I've decided to try Cauchy Criterion here. Consider $|x_m - x_n|$ for $m = 2n > n$:
$$
begin{align}
|x_m - x_n| &= left| 1+ sum_{k=1}^mfrac{k^2}{4^k} - 1 - sum_{k=1}^nfrac{k^2}{4^k}right|\
&= left|sum_{k=n+1}^{2n}frac{k^2}{4^k}right| \
&stackrel{>0}{=} sum_{k=n+1}^{2n}frac{k^2}{4^k} \
&le sum_{k=n+1}^{2n}frac{(2n)^2}{4^k} \
&le sum_{k=n+1}^{2n}frac{(2n)^2}{4^{n+1}} \
&= frac{4n^3}{4^{n+1}} \
&= frac{n^3}{4^n}
end{align}
$$
Consider the limit:
$$
0 le lim_{ntoinfty}|x_{2n} - x_n| le lim_{ntoinfty} frac{n^3}{4^n} = 0
$$
Which by squeeze theorem gives:
$$
lim_{ntoinfty}|x_{2n} - x_n| = 0
$$
proving $x_n$ is a fundamental sequence, hence convergent.
I've been also thinking of using monotone convergence theorem:
$$
x_{n+1} - x_n = frac{(n+1)^2}{4^{n+1}} > 0
$$
By this $x_n$ is monotonically increasing, but i couldn't find an upper bound.
I would like to know two things:
- It the above a rigorous way to show $x_n$ is convergent?
- Even though the problem statement only asks to prove convergence rather than find the limit I would still want to find it. W|A suggests its ${20over 27}$. Is it possible to obtain that result using more or less elementary calculus reasoning?(anything before the definition of a derivative)
calculus sequences-and-series limits proof-verification
asked Jan 9 at 15:44
romanroman
2,17321224
$endgroup$
2
$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:55
add a comment |
$begingroup$
Prove the following sequence is convergent:
$$
x_n = 1 + {1^2over 4} + {2^2over 4^2} + cdots + {n^2over 4^n}
$$
I've decided to try Cauchy Criterion here. Consider $|x_m - x_n|$ for $m = 2n > n$:
$$
begin{align}
|x_m - x_n| &= left| 1+ sum_{k=1}^mfrac{k^2}{4^k} - 1 - sum_{k=1}^nfrac{k^2}{4^k}right|\
&= left|sum_{k=n+1}^{2n}frac{k^2}{4^k}right| \
&stackrel{>0}{=} sum_{k=n+1}^{2n}frac{k^2}{4^k} \
&le sum_{k=n+1}^{2n}frac{(2n)^2}{4^k} \
&le sum_{k=n+1}^{2n}frac{(2n)^2}{4^{n+1}} \
&= frac{4n^3}{4^{n+1}} \
&= frac{n^3}{4^n}
end{align}
$$
Consider the limit:
$$
0 le lim_{ntoinfty}|x_{2n} - x_n| le lim_{ntoinfty} frac{n^3}{4^n} = 0
$$
Which by squeeze theorem gives:
$$
lim_{ntoinfty}|x_{2n} - x_n| = 0
$$
proving $x_n$ is a fundamental sequence, hence convergent.
I've been also thinking of using monotone convergence theorem:
$$
x_{n+1} - x_n = frac{(n+1)^2}{4^{n+1}} > 0
$$
By this $x_n$ is monotonically increasing, but i couldn't find an upper bound.
I would like to know two things:
- It the above a rigorous way to show $x_n$ is convergent?
- Even though the problem statement only asks to prove convergence rather than find the limit I would still want to find it. W|A suggests its ${20over 27}$. Is it possible to obtain that result using more or less elementary calculus reasoning?(anything before the definition of a derivative)
calculus sequences-and-series limits proof-verification
asked Jan 9 at 15:44
romanroman
2,17321224
$endgroup$
Prove the following sequence is convergent:
$$
x_n = 1 + {1^2over 4} + {2^2over 4^2} + cdots + {n^2over 4^n}
$$
I've decided to try Cauchy Criterion here. Consider $|x_m - x_n|$ for $m = 2n > n$:
$$
begin{align}
|x_m - x_n| &= left| 1+ sum_{k=1}^mfrac{k^2}{4^k} - 1 - sum_{k=1}^nfrac{k^2}{4^k}right|\
&= left|sum_{k=n+1}^{2n}frac{k^2}{4^k}right| \
&stackrel{>0}{=} sum_{k=n+1}^{2n}frac{k^2}{4^k} \
&le sum_{k=n+1}^{2n}frac{(2n)^2}{4^k} \
&le sum_{k=n+1}^{2n}frac{(2n)^2}{4^{n+1}} \
&= frac{4n^3}{4^{n+1}} \
&= frac{n^3}{4^n}
end{align}
$$
Consider the limit:
$$
0 le lim_{ntoinfty}|x_{2n} - x_n| le lim_{ntoinfty} frac{n^3}{4^n} = 0
$$
Which by squeeze theorem gives:
$$
lim_{ntoinfty}|x_{2n} - x_n| = 0
$$
proving $x_n$ is a fundamental sequence, hence convergent.
I've been also thinking of using monotone convergence theorem:
$$
x_{n+1} - x_n = frac{(n+1)^2}{4^{n+1}} > 0
$$
By this $x_n$ is monotonically increasing, but i couldn't find an upper bound.
I would like to know two things:
- It the above a rigorous way to show $x_n$ is convergent?
- Even though the problem statement only asks to prove convergence rather than find the limit I would still want to find it. W|A suggests its ${20over 27}$. Is it possible to obtain that result using more or less elementary calculus reasoning?(anything before the definition of a derivative)
calculus sequences-and-series limits proof-verification
calculus sequences-and-series limits proof-verification
asked Jan 9 at 15:44
romanroman
2,17321224
asked Jan 9 at 15:44
romanroman
2,17321224
asked Jan 9 at 15:44
romanroman
2,17321224
asked Jan 9 at 15:44
romanroman
2,17321224
asked Jan 9 at 15:44
romanroman
2,17321224
2,17321224
2
$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:55
add a comment |
2
$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:55
2
2
$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:55
$begingroup$
math.stackexchange.com/questions/593996/…
$endgroup$
– lab bhattacharjee
Jan 9 at 15:55
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint
For $x>4$ we have $$2^x>x^2$$therefore for $n>4$ we obtain $${n^2over 4^n}<{2^nover 4^n}={1over 2^n}$$
answered Jan 9 at 17:34
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
Using ratio test: $$lim_{ntoinfty}|frac{(n+1)^2 4^{-(n+1)}}{n^2 4^{-n}}|=frac{1}{4}<1$$ so the sequence of partial sums is convergent. Furthermore, by considering the geometric series $$frac{1}{1-x}=sum_{n=0}^infty x^n$$ $$frac{1}{(1-x)^2}=sum_{n=0}^infty nx^{n-1}$$ $$f(x)=frac{x}{(1-x)^2}=sum_{n=0}^infty nx^n$$
$$frac{f'(frac{1}{4})}{4}=sum_{n=0}^infty frac{n^2}{4^{n}}$$ you can easily deduce the limit of the sequence.
answered Jan 9 at 16:02
aledenaleden
2,057511
$endgroup$
1
$begingroup$
It is just adding a constant and does not depend on $n$, so it should not affect convergence.
$endgroup$
– aleden
Jan 9 at 16:13
add a comment |
$begingroup$
$frac{n^2}{4^n}=frac{1}{2^n}frac{n^2}{2^n}leq Cfrac{1}{2^n}$ for some constant $C,$ because $frac{n^2}{2^n}rightarrow 0$ and hence bounded sequence. So, $sum_n frac{n^2}{4^n}leq Cfrac{1}{2^n}<infty.$
answered Jan 9 at 15:55
John_WickJohn_Wick
1,621111
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
For $x>4$ we have $$2^x>x^2$$therefore for $n>4$ we obtain $${n^2over 4^n}<{2^nover 4^n}={1over 2^n}$$
answered Jan 9 at 17:34
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
Hint
For $x>4$ we have $$2^x>x^2$$therefore for $n>4$ we obtain $${n^2over 4^n}<{2^nover 4^n}={1over 2^n}$$
answered Jan 9 at 17:34
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
add a comment |
$begingroup$
Hint
For $x>4$ we have $$2^x>x^2$$therefore for $n>4$ we obtain $${n^2over 4^n}<{2^nover 4^n}={1over 2^n}$$
answered Jan 9 at 17:34
Mostafa AyazMostafa Ayaz
15.5k3939
$endgroup$
Hint
For $x>4$ we have $$2^x>x^2$$therefore for $n>4$ we obtain $${n^2over 4^n}<{2^nover 4^n}={1over 2^n}$$
answered Jan 9 at 17:34
Mostafa AyazMostafa Ayaz
15.5k3939
answered Jan 9 at 17:34
Mostafa AyazMostafa Ayaz
15.5k3939
answered Jan 9 at 17:34
Mostafa AyazMostafa Ayaz
15.5k3939
answered Jan 9 at 17:34
Mostafa AyazMostafa Ayaz
15.5k3939
15.5k3939
add a comment |
add a comment |
$begingroup$
Using ratio test: $$lim_{ntoinfty}|frac{(n+1)^2 4^{-(n+1)}}{n^2 4^{-n}}|=frac{1}{4}<1$$ so the sequence of partial sums is convergent. Furthermore, by considering the geometric series $$frac{1}{1-x}=sum_{n=0}^infty x^n$$ $$frac{1}{(1-x)^2}=sum_{n=0}^infty nx^{n-1}$$ $$f(x)=frac{x}{(1-x)^2}=sum_{n=0}^infty nx^n$$
$$frac{f'(frac{1}{4})}{4}=sum_{n=0}^infty frac{n^2}{4^{n}}$$ you can easily deduce the limit of the sequence.
answered Jan 9 at 16:02
aledenaleden
2,057511
$endgroup$
1
$begingroup$
It is just adding a constant and does not depend on $n$, so it should not affect convergence.
$endgroup$
– aleden
Jan 9 at 16:13
add a comment |
$begingroup$
Using ratio test: $$lim_{ntoinfty}|frac{(n+1)^2 4^{-(n+1)}}{n^2 4^{-n}}|=frac{1}{4}<1$$ so the sequence of partial sums is convergent. Furthermore, by considering the geometric series $$frac{1}{1-x}=sum_{n=0}^infty x^n$$ $$frac{1}{(1-x)^2}=sum_{n=0}^infty nx^{n-1}$$ $$f(x)=frac{x}{(1-x)^2}=sum_{n=0}^infty nx^n$$
$$frac{f'(frac{1}{4})}{4}=sum_{n=0}^infty frac{n^2}{4^{n}}$$ you can easily deduce the limit of the sequence.
answered Jan 9 at 16:02
aledenaleden
2,057511
$endgroup$
1
$begingroup$
It is just adding a constant and does not depend on $n$, so it should not affect convergence.
$endgroup$
– aleden
Jan 9 at 16:13
add a comment |
$begingroup$
Using ratio test: $$lim_{ntoinfty}|frac{(n+1)^2 4^{-(n+1)}}{n^2 4^{-n}}|=frac{1}{4}<1$$ so the sequence of partial sums is convergent. Furthermore, by considering the geometric series $$frac{1}{1-x}=sum_{n=0}^infty x^n$$ $$frac{1}{(1-x)^2}=sum_{n=0}^infty nx^{n-1}$$ $$f(x)=frac{x}{(1-x)^2}=sum_{n=0}^infty nx^n$$
$$frac{f'(frac{1}{4})}{4}=sum_{n=0}^infty frac{n^2}{4^{n}}$$ you can easily deduce the limit of the sequence.
answered Jan 9 at 16:02
aledenaleden
2,057511
$endgroup$
Using ratio test: $$lim_{ntoinfty}|frac{(n+1)^2 4^{-(n+1)}}{n^2 4^{-n}}|=frac{1}{4}<1$$ so the sequence of partial sums is convergent. Furthermore, by considering the geometric series $$frac{1}{1-x}=sum_{n=0}^infty x^n$$ $$frac{1}{(1-x)^2}=sum_{n=0}^infty nx^{n-1}$$ $$f(x)=frac{x}{(1-x)^2}=sum_{n=0}^infty nx^n$$
$$frac{f'(frac{1}{4})}{4}=sum_{n=0}^infty frac{n^2}{4^{n}}$$ you can easily deduce the limit of the sequence.
answered Jan 9 at 16:02
aledenaleden
2,057511
answered Jan 9 at 16:02
aledenaleden
2,057511
answered Jan 9 at 16:02
aledenaleden
2,057511
answered Jan 9 at 16:02
aledenaleden
2,057511
2,057511
1
$begingroup$
It is just adding a constant and does not depend on $n$, so it should not affect convergence.
$endgroup$
– aleden
Jan 9 at 16:13
add a comment |
1
$begingroup$
It is just adding a constant and does not depend on $n$, so it should not affect convergence.
$endgroup$
– aleden
Jan 9 at 16:13
1
1
$begingroup$
It is just adding a constant and does not depend on $n$, so it should not affect convergence.
$endgroup$
– aleden
Jan 9 at 16:13
$begingroup$
It is just adding a constant and does not depend on $n$, so it should not affect convergence.
$endgroup$
– aleden
Jan 9 at 16:13
add a comment |
$begingroup$
$frac{n^2}{4^n}=frac{1}{2^n}frac{n^2}{2^n}leq Cfrac{1}{2^n}$ for some constant $C,$ because $frac{n^2}{2^n}rightarrow 0$ and hence bounded sequence. So, $sum_n frac{n^2}{4^n}leq Cfrac{1}{2^n}<infty.$
answered Jan 9 at 15:55
John_WickJohn_Wick
1,621111
$endgroup$
add a comment |
$begingroup$
$frac{n^2}{4^n}=frac{1}{2^n}frac{n^2}{2^n}leq Cfrac{1}{2^n}$ for some constant $C,$ because $frac{n^2}{2^n}rightarrow 0$ and hence bounded sequence. So, $sum_n frac{n^2}{4^n}leq Cfrac{1}{2^n}<infty.$
answered Jan 9 at 15:55
John_WickJohn_Wick
1,621111
$endgroup$
add a comment |
$begingroup$
$frac{n^2}{4^n}=frac{1}{2^n}frac{n^2}{2^n}leq Cfrac{1}{2^n}$ for some constant $C,$ because $frac{n^2}{2^n}rightarrow 0$ and hence bounded sequence. So, $sum_n frac{n^2}{4^n}leq Cfrac{1}{2^n}<infty.$
answered Jan 9 at 15:55
John_WickJohn_Wick
1,621111
$endgroup$
$frac{n^2}{4^n}=frac{1}{2^n}frac{n^2}{2^n}leq Cfrac{1}{2^n}$ for some constant $C,$ because $frac{n^2}{2^n}rightarrow 0$ and hence bounded sequence. So, $sum_n frac{n^2}{4^n}leq Cfrac{1}{2^n}<infty.$
answered Jan 9 at 15:55
John_WickJohn_Wick
1,621111
answered Jan 9 at 15:55
John_WickJohn_Wick
1,621111
answered Jan 9 at 15:55
John_WickJohn_Wick
1,621111
answered Jan 9 at 15:55
John_WickJohn_Wick
1,621111
1,621111
add a comment |
add a comment |
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– lab bhattacharjee
Jan 9 at 15:55