Why is the divergence of this series apparently not predicted by the Monotonic Sequence Theorem?Prove if the...
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Why is the divergence of this series apparently not predicted by the Monotonic Sequence Theorem?
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As far as I understand, the Monotonic Sequence Theorem states that if a sequence is monotonic and the individual terms are bounded, then the sequence is convergent.
My book states that $lim limits_{t to infty}sum_{n=1}^t b^{ln n}$ is convergent only for b $lt$ $frac{1}{e}$. However, is $lim limits_{t to infty}sum_{n=1}^t 0.5^{ln n}$, for example, not a monotonically decreasing series, whose terms are bound by 0 below and 1 above? Therefore this series meets the criteria for the MST, yet diverges, and does not share the outcome predicted by the theorem.
Why is $lim limits_{t to infty}sum_{n=1}^t 0.5^{ln n}$ divergent when it appears to meet the Monotonic Sequence Theorem's criteria for convergence?
calculus sequences-and-series convergence
asked 8 hours ago
sirmax224sirmax224
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As far as I understand, the Monotonic Sequence Theorem states that if a sequence is monotonic and the individual terms are bounded, then the sequence is convergent.
My book states that $lim limits_{t to infty}sum_{n=1}^t b^{ln n}$ is convergent only for b $lt$ $frac{1}{e}$. However, is $lim limits_{t to infty}sum_{n=1}^t 0.5^{ln n}$, for example, not a monotonically decreasing series, whose terms are bound by 0 below and 1 above? Therefore this series meets the criteria for the MST, yet diverges, and does not share the outcome predicted by the theorem.
Why is $lim limits_{t to infty}sum_{n=1}^t 0.5^{ln n}$ divergent when it appears to meet the Monotonic Sequence Theorem's criteria for convergence?
calculus sequences-and-series convergence
asked 8 hours ago
sirmax224sirmax224
183 bronze badges
New contributor
sirmax224 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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3
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There is a difference between the sequence and the series. The sequence ${1 over n}$ is bounded and monotonic and has a limit, but $sum_k {1 over k}$ does not.
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– copper.hat
8 hours ago
add a comment |
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As far as I understand, the Monotonic Sequence Theorem states that if a sequence is monotonic and the individual terms are bounded, then the sequence is convergent.
My book states that $lim limits_{t to infty}sum_{n=1}^t b^{ln n}$ is convergent only for b $lt$ $frac{1}{e}$. However, is $lim limits_{t to infty}sum_{n=1}^t 0.5^{ln n}$, for example, not a monotonically decreasing series, whose terms are bound by 0 below and 1 above? Therefore this series meets the criteria for the MST, yet diverges, and does not share the outcome predicted by the theorem.
Why is $lim limits_{t to infty}sum_{n=1}^t 0.5^{ln n}$ divergent when it appears to meet the Monotonic Sequence Theorem's criteria for convergence?
calculus sequences-and-series convergence
asked 8 hours ago
sirmax224sirmax224
183 bronze badges
New contributor
sirmax224 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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As far as I understand, the Monotonic Sequence Theorem states that if a sequence is monotonic and the individual terms are bounded, then the sequence is convergent.
My book states that $lim limits_{t to infty}sum_{n=1}^t b^{ln n}$ is convergent only for b $lt$ $frac{1}{e}$. However, is $lim limits_{t to infty}sum_{n=1}^t 0.5^{ln n}$, for example, not a monotonically decreasing series, whose terms are bound by 0 below and 1 above? Therefore this series meets the criteria for the MST, yet diverges, and does not share the outcome predicted by the theorem.
Why is $lim limits_{t to infty}sum_{n=1}^t 0.5^{ln n}$ divergent when it appears to meet the Monotonic Sequence Theorem's criteria for convergence?
calculus sequences-and-series convergence
calculus sequences-and-series convergence
asked 8 hours ago
sirmax224sirmax224
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sirmax224 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 8 hours ago
sirmax224sirmax224
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asked 8 hours ago
sirmax224sirmax224
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asked 8 hours ago
sirmax224sirmax224
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asked 8 hours ago
sirmax224sirmax224
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183 bronze badges
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3
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There is a difference between the sequence and the series. The sequence ${1 over n}$ is bounded and monotonic and has a limit, but $sum_k {1 over k}$ does not.
$endgroup$
– copper.hat
8 hours ago
add a comment |
3
$begingroup$
There is a difference between the sequence and the series. The sequence ${1 over n}$ is bounded and monotonic and has a limit, but $sum_k {1 over k}$ does not.
$endgroup$
– copper.hat
8 hours ago
3
3
$begingroup$
There is a difference between the sequence and the series. The sequence ${1 over n}$ is bounded and monotonic and has a limit, but $sum_k {1 over k}$ does not.
$endgroup$
– copper.hat
8 hours ago
$begingroup$
There is a difference between the sequence and the series. The sequence ${1 over n}$ is bounded and monotonic and has a limit, but $sum_k {1 over k}$ does not.
$endgroup$
– copper.hat
8 hours ago
add a comment |
2 Answers
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You've confused sequence convergence (the $n$th term has a finite $ntoinfty$ limit) with series convergence (the sum of the first $n$ terms has a finite $ntoinfty$ limit).
answered 8 hours ago
J.G.J.G.
40.8k2 gold badges38 silver badges59 bronze badges
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$$lim_{ttoinfty} sum_{n=1}^infty b^{ln n}=sum_{n=1}^infty b^{ln n}=sum_{n=1}^infty e^{ln bln n}=sum_{n=1}^infty n^{ln b}$$
This series converges iff $ln b<-1$,i.e. when $b<1/e$.
Now, $0.5>1/e$. So, you certainly have that when $b=0.5$ this diverges. This doesn't contradict monotonic sequence theorem, even though $ (0.5)^{ln n}$ is bounded by $1$ for all $n$, you don't have that $sum_{n=1}^infty (0.5)^{ln n}$ is bounded for all $t$.
answered 7 hours ago
Julian MejiaJulian Mejia
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2 Answers
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You've confused sequence convergence (the $n$th term has a finite $ntoinfty$ limit) with series convergence (the sum of the first $n$ terms has a finite $ntoinfty$ limit).
answered 8 hours ago
J.G.J.G.
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You've confused sequence convergence (the $n$th term has a finite $ntoinfty$ limit) with series convergence (the sum of the first $n$ terms has a finite $ntoinfty$ limit).
answered 8 hours ago
J.G.J.G.
40.8k2 gold badges38 silver badges59 bronze badges
$endgroup$
add a comment |
$begingroup$
You've confused sequence convergence (the $n$th term has a finite $ntoinfty$ limit) with series convergence (the sum of the first $n$ terms has a finite $ntoinfty$ limit).
answered 8 hours ago
J.G.J.G.
40.8k2 gold badges38 silver badges59 bronze badges
$endgroup$
You've confused sequence convergence (the $n$th term has a finite $ntoinfty$ limit) with series convergence (the sum of the first $n$ terms has a finite $ntoinfty$ limit).
answered 8 hours ago
J.G.J.G.
40.8k2 gold badges38 silver badges59 bronze badges
answered 8 hours ago
J.G.J.G.
40.8k2 gold badges38 silver badges59 bronze badges
answered 8 hours ago
J.G.J.G.
40.8k2 gold badges38 silver badges59 bronze badges
answered 8 hours ago
J.G.J.G.
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40.8k2 gold badges38 silver badges59 bronze badges
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$$lim_{ttoinfty} sum_{n=1}^infty b^{ln n}=sum_{n=1}^infty b^{ln n}=sum_{n=1}^infty e^{ln bln n}=sum_{n=1}^infty n^{ln b}$$
This series converges iff $ln b<-1$,i.e. when $b<1/e$.
Now, $0.5>1/e$. So, you certainly have that when $b=0.5$ this diverges. This doesn't contradict monotonic sequence theorem, even though $ (0.5)^{ln n}$ is bounded by $1$ for all $n$, you don't have that $sum_{n=1}^infty (0.5)^{ln n}$ is bounded for all $t$.
answered 7 hours ago
Julian MejiaJulian Mejia
3,7254 silver badges16 bronze badges
$endgroup$
add a comment |
$begingroup$
$$lim_{ttoinfty} sum_{n=1}^infty b^{ln n}=sum_{n=1}^infty b^{ln n}=sum_{n=1}^infty e^{ln bln n}=sum_{n=1}^infty n^{ln b}$$
This series converges iff $ln b<-1$,i.e. when $b<1/e$.
Now, $0.5>1/e$. So, you certainly have that when $b=0.5$ this diverges. This doesn't contradict monotonic sequence theorem, even though $ (0.5)^{ln n}$ is bounded by $1$ for all $n$, you don't have that $sum_{n=1}^infty (0.5)^{ln n}$ is bounded for all $t$.
answered 7 hours ago
Julian MejiaJulian Mejia
3,7254 silver badges16 bronze badges
$endgroup$
add a comment |
$begingroup$
$$lim_{ttoinfty} sum_{n=1}^infty b^{ln n}=sum_{n=1}^infty b^{ln n}=sum_{n=1}^infty e^{ln bln n}=sum_{n=1}^infty n^{ln b}$$
This series converges iff $ln b<-1$,i.e. when $b<1/e$.
Now, $0.5>1/e$. So, you certainly have that when $b=0.5$ this diverges. This doesn't contradict monotonic sequence theorem, even though $ (0.5)^{ln n}$ is bounded by $1$ for all $n$, you don't have that $sum_{n=1}^infty (0.5)^{ln n}$ is bounded for all $t$.
answered 7 hours ago
Julian MejiaJulian Mejia
3,7254 silver badges16 bronze badges
$endgroup$
$$lim_{ttoinfty} sum_{n=1}^infty b^{ln n}=sum_{n=1}^infty b^{ln n}=sum_{n=1}^infty e^{ln bln n}=sum_{n=1}^infty n^{ln b}$$
This series converges iff $ln b<-1$,i.e. when $b<1/e$.
Now, $0.5>1/e$. So, you certainly have that when $b=0.5$ this diverges. This doesn't contradict monotonic sequence theorem, even though $ (0.5)^{ln n}$ is bounded by $1$ for all $n$, you don't have that $sum_{n=1}^infty (0.5)^{ln n}$ is bounded for all $t$.
answered 7 hours ago
Julian MejiaJulian Mejia
3,7254 silver badges16 bronze badges
answered 7 hours ago
Julian MejiaJulian Mejia
3,7254 silver badges16 bronze badges
answered 7 hours ago
Julian MejiaJulian Mejia
3,7254 silver badges16 bronze badges
answered 7 hours ago
Julian MejiaJulian Mejia
3,7254 silver badges16 bronze badges
3,7254 silver badges16 bronze badges
add a comment |
add a comment |
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There is a difference between the sequence and the series. The sequence ${1 over n}$ is bounded and monotonic and has a limit, but $sum_k {1 over k}$ does not.
$endgroup$
– copper.hat
8 hours ago