Mixing
Does $sum q_i = infty$ imply $sum log(1+q_i)=infty$?
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My intuition is that the first-order term in the Taylor expansion should dominate the series, if divergent:
$$
log(1+x) = x - frac { x ^ { 2 } } { 2 } + frac { x ^ { 3 } } { 3 } - frac { x ^ { 4 } } { 4 } + cdots
$$
So we would get with Taylor expansion:
$$
sum_ilog(1+q_i) = sum_{i}sum_{k=1}^inftyfrac{q_i^k}{k} = sum_i q_i + sum_{i}sum_{k=2}^inftyfrac{q_i^k}{k}
$$
Is there a slick way to control the residual?
power-series taylor-expansion divergent-series
asked Jan 10 at 18:24
nakajuicenakajuice
1,37811323
$endgroup$
add a comment |
$begingroup$
My intuition is that the first-order term in the Taylor expansion should dominate the series, if divergent:
$$
log(1+x) = x - frac { x ^ { 2 } } { 2 } + frac { x ^ { 3 } } { 3 } - frac { x ^ { 4 } } { 4 } + cdots
$$
So we would get with Taylor expansion:
$$
sum_ilog(1+q_i) = sum_{i}sum_{k=1}^inftyfrac{q_i^k}{k} = sum_i q_i + sum_{i}sum_{k=2}^inftyfrac{q_i^k}{k}
$$
Is there a slick way to control the residual?
power-series taylor-expansion divergent-series
asked Jan 10 at 18:24
nakajuicenakajuice
1,37811323
$endgroup$
$begingroup$
If $q_i$'s are all non-negative, then the answer is yes. Otherwise, we have counter-examples.
$endgroup$
– Sangchul Lee
Jan 10 at 18:28
$begingroup$
Can you please come up with one, very curious.
$endgroup$
– nakajuice
Jan 10 at 18:42
$begingroup$
I will post an explicit construction when I am sit at my PC. But the idea is that, if $r_i = log(1+q_i)$ so that $q_i = e^{r_i}-1$, then $q_i > r_i$ unless $r_i=0$. So, with a careful choice of $r_i$'s, you can make $sum_i(q_i-r_i)=infty$ while $sum_i r_i$ converge.
$endgroup$
– Sangchul Lee
Jan 10 at 18:48
$begingroup$
@MathLover nice, thank you
$endgroup$
– nakajuice
Jan 10 at 18:56
add a comment |
$begingroup$
My intuition is that the first-order term in the Taylor expansion should dominate the series, if divergent:
$$
log(1+x) = x - frac { x ^ { 2 } } { 2 } + frac { x ^ { 3 } } { 3 } - frac { x ^ { 4 } } { 4 } + cdots
$$
So we would get with Taylor expansion:
$$
sum_ilog(1+q_i) = sum_{i}sum_{k=1}^inftyfrac{q_i^k}{k} = sum_i q_i + sum_{i}sum_{k=2}^inftyfrac{q_i^k}{k}
$$
Is there a slick way to control the residual?
power-series taylor-expansion divergent-series
asked Jan 10 at 18:24
nakajuicenakajuice
1,37811323
$endgroup$
My intuition is that the first-order term in the Taylor expansion should dominate the series, if divergent:
$$
log(1+x) = x - frac { x ^ { 2 } } { 2 } + frac { x ^ { 3 } } { 3 } - frac { x ^ { 4 } } { 4 } + cdots
$$
So we would get with Taylor expansion:
$$
sum_ilog(1+q_i) = sum_{i}sum_{k=1}^inftyfrac{q_i^k}{k} = sum_i q_i + sum_{i}sum_{k=2}^inftyfrac{q_i^k}{k}
$$
Is there a slick way to control the residual?
power-series taylor-expansion divergent-series
power-series taylor-expansion divergent-series
asked Jan 10 at 18:24
nakajuicenakajuice
1,37811323
asked Jan 10 at 18:24
nakajuicenakajuice
1,37811323
asked Jan 10 at 18:24
nakajuicenakajuice
1,37811323
asked Jan 10 at 18:24
nakajuicenakajuice
1,37811323
asked Jan 10 at 18:24
nakajuicenakajuice
1,37811323
1,37811323
$begingroup$
If $q_i$'s are all non-negative, then the answer is yes. Otherwise, we have counter-examples.
$endgroup$
– Sangchul Lee
Jan 10 at 18:28
$begingroup$
Can you please come up with one, very curious.
$endgroup$
– nakajuice
Jan 10 at 18:42
$begingroup$
I will post an explicit construction when I am sit at my PC. But the idea is that, if $r_i = log(1+q_i)$ so that $q_i = e^{r_i}-1$, then $q_i > r_i$ unless $r_i=0$. So, with a careful choice of $r_i$'s, you can make $sum_i(q_i-r_i)=infty$ while $sum_i r_i$ converge.
$endgroup$
– Sangchul Lee
Jan 10 at 18:48
$begingroup$
@MathLover nice, thank you
$endgroup$
– nakajuice
Jan 10 at 18:56
add a comment |
$begingroup$
If $q_i$'s are all non-negative, then the answer is yes. Otherwise, we have counter-examples.
$endgroup$
– Sangchul Lee
Jan 10 at 18:28
$begingroup$
Can you please come up with one, very curious.
$endgroup$
– nakajuice
Jan 10 at 18:42
$begingroup$
I will post an explicit construction when I am sit at my PC. But the idea is that, if $r_i = log(1+q_i)$ so that $q_i = e^{r_i}-1$, then $q_i > r_i$ unless $r_i=0$. So, with a careful choice of $r_i$'s, you can make $sum_i(q_i-r_i)=infty$ while $sum_i r_i$ converge.
$endgroup$
– Sangchul Lee
Jan 10 at 18:48
$begingroup$
@MathLover nice, thank you
$endgroup$
– nakajuice
Jan 10 at 18:56
$begingroup$
If $q_i$'s are all non-negative, then the answer is yes. Otherwise, we have counter-examples.
$endgroup$
– Sangchul Lee
Jan 10 at 18:28
$begingroup$
If $q_i$'s are all non-negative, then the answer is yes. Otherwise, we have counter-examples.
$endgroup$
– Sangchul Lee
Jan 10 at 18:28
$begingroup$
Can you please come up with one, very curious.
$endgroup$
– nakajuice
Jan 10 at 18:42
$begingroup$
Can you please come up with one, very curious.
$endgroup$
– nakajuice
Jan 10 at 18:42
$begingroup$
I will post an explicit construction when I am sit at my PC. But the idea is that, if $r_i = log(1+q_i)$ so that $q_i = e^{r_i}-1$, then $q_i > r_i$ unless $r_i=0$. So, with a careful choice of $r_i$'s, you can make $sum_i(q_i-r_i)=infty$ while $sum_i r_i$ converge.
$endgroup$
– Sangchul Lee
Jan 10 at 18:48
$begingroup$
I will post an explicit construction when I am sit at my PC. But the idea is that, if $r_i = log(1+q_i)$ so that $q_i = e^{r_i}-1$, then $q_i > r_i$ unless $r_i=0$. So, with a careful choice of $r_i$'s, you can make $sum_i(q_i-r_i)=infty$ while $sum_i r_i$ converge.
$endgroup$
– Sangchul Lee
Jan 10 at 18:48
$begingroup$
@MathLover nice, thank you
$endgroup$
– nakajuice
Jan 10 at 18:56
$begingroup$
@MathLover nice, thank you
$endgroup$
– nakajuice
Jan 10 at 18:56
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint:
Consider $q_0=1$, $q_1=-1/2$, and $q_{i+2}=q_i$ for $i=0,1,2,cdots$
answered Jan 10 at 18:59
Math LoverMath Lover
14k31436
$endgroup$
add a comment |
$begingroup$
For $q_i>0$ or equivalently $ln (1+q_i)>0$, each series fails to equal $infty$ iff it has a finite limit instead. This is where it helps to work with the contrapositive. If $sum_iln (1+q_i)$ is finite for an infinite sequence $q_i$, $lim_{itoinfty}ln (1+q_i)=0$ so $lim_{itoinfty}q_i=0$. Since $ln (1+q_i)sim q_i$ for small $q_i$, this implies $sum_i q_i$ also converges. A comment of Sangchui Lee's references the fact that, with a suitable choice of not-all-positive $q_i$ (viz MathLover's answer), the sum of logarithms might approach neither $infty$ nor $-infty$. (Their example gives partial sums of the $ln (1+q_i)$ equal to either $ln 2$ or $0$, so there's no limit, infinite or otherwise.)
answered Jan 10 at 18:42
J.G.J.G.
25.6k22539
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add a comment |
$begingroup$
User @Math Lover provided an example where $ sum_{n=0}^{infty} q_n = infty$ but $ sum_{n=0}^{N} log(1+q_n)$ alternates between $log 2$ and $0$ in $N$. In this answer, we construct an example where
$$ sum_{n=0}^{infty} q_n = infty quad text{and} quad sum_{n=0}^{infty} log(1+q_n) = 0.$$
To this end, we prepare two auxiliary sequences:
$(epsilon_k)_{k=1}^{infty}$ is a sequence such that $1 geq epsilon_k geq epsilon_{k+1} > 0$ and $epsilon_k to 0$ as $k to infty$.
$(N_k)_{k=1}^{infty}$ is a sequence of positive integers such that $ N_k epsilon_k^2 geq 1$ for all $k$.
Now define $(r_n)$ and $(q_n)$ by
$$ (r_n)_{n=0}^{infty} = ( underbrace{ epsilon_1, -epsilon_1, cdots, epsilon_1, -epsilon_1 }_{2N_1text{-terms}}, underbrace{ epsilon_2, -epsilon_2, cdots, epsilon_2, -epsilon_2 }_{2N_2text{-terms}}, cdots ),
qquad q_n = e^{r_n} - 1. $$
Then $r_n = log(1+q_n)$, and
By the alternating series test, $sum_{n=0}^{infty} r_n$ converges. Moreover, its odd-th partial sums are identically zero, hence the sum is also zero.
Using the fact that $e^x geq 1 + x + frac{1}{e}x^2$ for $|x| leq 1$, it follows that
$$ q_n geq r_n + frac{1}{e}r_n^2 $$
Now summing over $n$, we obtain
$$ sum_{n=0}^{infty} q_n geq left( sum_{n=0}^{infty} r_n right) + left( sum_{k=1}^{infty} 2N_k cdot frac{epsilon_k^2}{e} right) geq sum_{k=1}^{infty} frac{2}{e} = infty. $$
answered Jan 11 at 2:05
Sangchul LeeSangchul Lee
93k12167269
$endgroup$
add a comment |
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3 Answers
3
active
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3 Answers
3
active
oldest
votes
active
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oldest
votes
$begingroup$
Hint:
Consider $q_0=1$, $q_1=-1/2$, and $q_{i+2}=q_i$ for $i=0,1,2,cdots$
answered Jan 10 at 18:59
Math LoverMath Lover
14k31436
$endgroup$
add a comment |
$begingroup$
Hint:
Consider $q_0=1$, $q_1=-1/2$, and $q_{i+2}=q_i$ for $i=0,1,2,cdots$
answered Jan 10 at 18:59
Math LoverMath Lover
14k31436
$endgroup$
add a comment |
$begingroup$
Hint:
Consider $q_0=1$, $q_1=-1/2$, and $q_{i+2}=q_i$ for $i=0,1,2,cdots$
answered Jan 10 at 18:59
Math LoverMath Lover
14k31436
$endgroup$
Hint:
Consider $q_0=1$, $q_1=-1/2$, and $q_{i+2}=q_i$ for $i=0,1,2,cdots$
answered Jan 10 at 18:59
Math LoverMath Lover
14k31436
answered Jan 10 at 18:59
Math LoverMath Lover
14k31436
answered Jan 10 at 18:59
Math LoverMath Lover
14k31436
answered Jan 10 at 18:59
Math LoverMath Lover
14k31436
14k31436
add a comment |
add a comment |
$begingroup$
For $q_i>0$ or equivalently $ln (1+q_i)>0$, each series fails to equal $infty$ iff it has a finite limit instead. This is where it helps to work with the contrapositive. If $sum_iln (1+q_i)$ is finite for an infinite sequence $q_i$, $lim_{itoinfty}ln (1+q_i)=0$ so $lim_{itoinfty}q_i=0$. Since $ln (1+q_i)sim q_i$ for small $q_i$, this implies $sum_i q_i$ also converges. A comment of Sangchui Lee's references the fact that, with a suitable choice of not-all-positive $q_i$ (viz MathLover's answer), the sum of logarithms might approach neither $infty$ nor $-infty$. (Their example gives partial sums of the $ln (1+q_i)$ equal to either $ln 2$ or $0$, so there's no limit, infinite or otherwise.)
answered Jan 10 at 18:42
J.G.J.G.
25.6k22539
$endgroup$
add a comment |
$begingroup$
For $q_i>0$ or equivalently $ln (1+q_i)>0$, each series fails to equal $infty$ iff it has a finite limit instead. This is where it helps to work with the contrapositive. If $sum_iln (1+q_i)$ is finite for an infinite sequence $q_i$, $lim_{itoinfty}ln (1+q_i)=0$ so $lim_{itoinfty}q_i=0$. Since $ln (1+q_i)sim q_i$ for small $q_i$, this implies $sum_i q_i$ also converges. A comment of Sangchui Lee's references the fact that, with a suitable choice of not-all-positive $q_i$ (viz MathLover's answer), the sum of logarithms might approach neither $infty$ nor $-infty$. (Their example gives partial sums of the $ln (1+q_i)$ equal to either $ln 2$ or $0$, so there's no limit, infinite or otherwise.)
answered Jan 10 at 18:42
J.G.J.G.
25.6k22539
$endgroup$
add a comment |
$begingroup$
For $q_i>0$ or equivalently $ln (1+q_i)>0$, each series fails to equal $infty$ iff it has a finite limit instead. This is where it helps to work with the contrapositive. If $sum_iln (1+q_i)$ is finite for an infinite sequence $q_i$, $lim_{itoinfty}ln (1+q_i)=0$ so $lim_{itoinfty}q_i=0$. Since $ln (1+q_i)sim q_i$ for small $q_i$, this implies $sum_i q_i$ also converges. A comment of Sangchui Lee's references the fact that, with a suitable choice of not-all-positive $q_i$ (viz MathLover's answer), the sum of logarithms might approach neither $infty$ nor $-infty$. (Their example gives partial sums of the $ln (1+q_i)$ equal to either $ln 2$ or $0$, so there's no limit, infinite or otherwise.)
answered Jan 10 at 18:42
J.G.J.G.
25.6k22539
$endgroup$
For $q_i>0$ or equivalently $ln (1+q_i)>0$, each series fails to equal $infty$ iff it has a finite limit instead. This is where it helps to work with the contrapositive. If $sum_iln (1+q_i)$ is finite for an infinite sequence $q_i$, $lim_{itoinfty}ln (1+q_i)=0$ so $lim_{itoinfty}q_i=0$. Since $ln (1+q_i)sim q_i$ for small $q_i$, this implies $sum_i q_i$ also converges. A comment of Sangchui Lee's references the fact that, with a suitable choice of not-all-positive $q_i$ (viz MathLover's answer), the sum of logarithms might approach neither $infty$ nor $-infty$. (Their example gives partial sums of the $ln (1+q_i)$ equal to either $ln 2$ or $0$, so there's no limit, infinite or otherwise.)
answered Jan 10 at 18:42
J.G.J.G.
25.6k22539
edited Jan 10 at 19:23
answered Jan 10 at 18:42
J.G.J.G.
25.6k22539
answered Jan 10 at 18:42
J.G.J.G.
25.6k22539
answered Jan 10 at 18:42
J.G.J.G.
25.6k22539
25.6k22539
add a comment |
add a comment |
$begingroup$
User @Math Lover provided an example where $ sum_{n=0}^{infty} q_n = infty$ but $ sum_{n=0}^{N} log(1+q_n)$ alternates between $log 2$ and $0$ in $N$. In this answer, we construct an example where
$$ sum_{n=0}^{infty} q_n = infty quad text{and} quad sum_{n=0}^{infty} log(1+q_n) = 0.$$
To this end, we prepare two auxiliary sequences:
$(epsilon_k)_{k=1}^{infty}$ is a sequence such that $1 geq epsilon_k geq epsilon_{k+1} > 0$ and $epsilon_k to 0$ as $k to infty$.
$(N_k)_{k=1}^{infty}$ is a sequence of positive integers such that $ N_k epsilon_k^2 geq 1$ for all $k$.
Now define $(r_n)$ and $(q_n)$ by
$$ (r_n)_{n=0}^{infty} = ( underbrace{ epsilon_1, -epsilon_1, cdots, epsilon_1, -epsilon_1 }_{2N_1text{-terms}}, underbrace{ epsilon_2, -epsilon_2, cdots, epsilon_2, -epsilon_2 }_{2N_2text{-terms}}, cdots ),
qquad q_n = e^{r_n} - 1. $$
Then $r_n = log(1+q_n)$, and
By the alternating series test, $sum_{n=0}^{infty} r_n$ converges. Moreover, its odd-th partial sums are identically zero, hence the sum is also zero.
Using the fact that $e^x geq 1 + x + frac{1}{e}x^2$ for $|x| leq 1$, it follows that
$$ q_n geq r_n + frac{1}{e}r_n^2 $$
Now summing over $n$, we obtain
$$ sum_{n=0}^{infty} q_n geq left( sum_{n=0}^{infty} r_n right) + left( sum_{k=1}^{infty} 2N_k cdot frac{epsilon_k^2}{e} right) geq sum_{k=1}^{infty} frac{2}{e} = infty. $$
answered Jan 11 at 2:05
Sangchul LeeSangchul Lee
93k12167269
$endgroup$
add a comment |
$begingroup$
User @Math Lover provided an example where $ sum_{n=0}^{infty} q_n = infty$ but $ sum_{n=0}^{N} log(1+q_n)$ alternates between $log 2$ and $0$ in $N$. In this answer, we construct an example where
$$ sum_{n=0}^{infty} q_n = infty quad text{and} quad sum_{n=0}^{infty} log(1+q_n) = 0.$$
To this end, we prepare two auxiliary sequences:
$(epsilon_k)_{k=1}^{infty}$ is a sequence such that $1 geq epsilon_k geq epsilon_{k+1} > 0$ and $epsilon_k to 0$ as $k to infty$.
$(N_k)_{k=1}^{infty}$ is a sequence of positive integers such that $ N_k epsilon_k^2 geq 1$ for all $k$.
Now define $(r_n)$ and $(q_n)$ by
$$ (r_n)_{n=0}^{infty} = ( underbrace{ epsilon_1, -epsilon_1, cdots, epsilon_1, -epsilon_1 }_{2N_1text{-terms}}, underbrace{ epsilon_2, -epsilon_2, cdots, epsilon_2, -epsilon_2 }_{2N_2text{-terms}}, cdots ),
qquad q_n = e^{r_n} - 1. $$
Then $r_n = log(1+q_n)$, and
By the alternating series test, $sum_{n=0}^{infty} r_n$ converges. Moreover, its odd-th partial sums are identically zero, hence the sum is also zero.
Using the fact that $e^x geq 1 + x + frac{1}{e}x^2$ for $|x| leq 1$, it follows that
$$ q_n geq r_n + frac{1}{e}r_n^2 $$
Now summing over $n$, we obtain
$$ sum_{n=0}^{infty} q_n geq left( sum_{n=0}^{infty} r_n right) + left( sum_{k=1}^{infty} 2N_k cdot frac{epsilon_k^2}{e} right) geq sum_{k=1}^{infty} frac{2}{e} = infty. $$
answered Jan 11 at 2:05
Sangchul LeeSangchul Lee
93k12167269
$endgroup$
add a comment |
$begingroup$
User @Math Lover provided an example where $ sum_{n=0}^{infty} q_n = infty$ but $ sum_{n=0}^{N} log(1+q_n)$ alternates between $log 2$ and $0$ in $N$. In this answer, we construct an example where
$$ sum_{n=0}^{infty} q_n = infty quad text{and} quad sum_{n=0}^{infty} log(1+q_n) = 0.$$
To this end, we prepare two auxiliary sequences:
$(epsilon_k)_{k=1}^{infty}$ is a sequence such that $1 geq epsilon_k geq epsilon_{k+1} > 0$ and $epsilon_k to 0$ as $k to infty$.
$(N_k)_{k=1}^{infty}$ is a sequence of positive integers such that $ N_k epsilon_k^2 geq 1$ for all $k$.
Now define $(r_n)$ and $(q_n)$ by
$$ (r_n)_{n=0}^{infty} = ( underbrace{ epsilon_1, -epsilon_1, cdots, epsilon_1, -epsilon_1 }_{2N_1text{-terms}}, underbrace{ epsilon_2, -epsilon_2, cdots, epsilon_2, -epsilon_2 }_{2N_2text{-terms}}, cdots ),
qquad q_n = e^{r_n} - 1. $$
Then $r_n = log(1+q_n)$, and
By the alternating series test, $sum_{n=0}^{infty} r_n$ converges. Moreover, its odd-th partial sums are identically zero, hence the sum is also zero.
Using the fact that $e^x geq 1 + x + frac{1}{e}x^2$ for $|x| leq 1$, it follows that
$$ q_n geq r_n + frac{1}{e}r_n^2 $$
Now summing over $n$, we obtain
$$ sum_{n=0}^{infty} q_n geq left( sum_{n=0}^{infty} r_n right) + left( sum_{k=1}^{infty} 2N_k cdot frac{epsilon_k^2}{e} right) geq sum_{k=1}^{infty} frac{2}{e} = infty. $$
answered Jan 11 at 2:05
Sangchul LeeSangchul Lee
93k12167269
$endgroup$
User @Math Lover provided an example where $ sum_{n=0}^{infty} q_n = infty$ but $ sum_{n=0}^{N} log(1+q_n)$ alternates between $log 2$ and $0$ in $N$. In this answer, we construct an example where
$$ sum_{n=0}^{infty} q_n = infty quad text{and} quad sum_{n=0}^{infty} log(1+q_n) = 0.$$
To this end, we prepare two auxiliary sequences:
$(epsilon_k)_{k=1}^{infty}$ is a sequence such that $1 geq epsilon_k geq epsilon_{k+1} > 0$ and $epsilon_k to 0$ as $k to infty$.
$(N_k)_{k=1}^{infty}$ is a sequence of positive integers such that $ N_k epsilon_k^2 geq 1$ for all $k$.
Now define $(r_n)$ and $(q_n)$ by
$$ (r_n)_{n=0}^{infty} = ( underbrace{ epsilon_1, -epsilon_1, cdots, epsilon_1, -epsilon_1 }_{2N_1text{-terms}}, underbrace{ epsilon_2, -epsilon_2, cdots, epsilon_2, -epsilon_2 }_{2N_2text{-terms}}, cdots ),
qquad q_n = e^{r_n} - 1. $$
Then $r_n = log(1+q_n)$, and
By the alternating series test, $sum_{n=0}^{infty} r_n$ converges. Moreover, its odd-th partial sums are identically zero, hence the sum is also zero.
Using the fact that $e^x geq 1 + x + frac{1}{e}x^2$ for $|x| leq 1$, it follows that
$$ q_n geq r_n + frac{1}{e}r_n^2 $$
Now summing over $n$, we obtain
$$ sum_{n=0}^{infty} q_n geq left( sum_{n=0}^{infty} r_n right) + left( sum_{k=1}^{infty} 2N_k cdot frac{epsilon_k^2}{e} right) geq sum_{k=1}^{infty} frac{2}{e} = infty. $$
answered Jan 11 at 2:05
Sangchul LeeSangchul Lee
93k12167269
answered Jan 11 at 2:05
Sangchul LeeSangchul Lee
93k12167269
answered Jan 11 at 2:05
Sangchul LeeSangchul Lee
93k12167269
answered Jan 11 at 2:05
Sangchul LeeSangchul Lee
93k12167269
93k12167269
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$begingroup$
If $q_i$'s are all non-negative, then the answer is yes. Otherwise, we have counter-examples.
$endgroup$
– Sangchul Lee
Jan 10 at 18:28
$begingroup$
Can you please come up with one, very curious.
$endgroup$
– nakajuice
Jan 10 at 18:42
$begingroup$
I will post an explicit construction when I am sit at my PC. But the idea is that, if $r_i = log(1+q_i)$ so that $q_i = e^{r_i}-1$, then $q_i > r_i$ unless $r_i=0$. So, with a careful choice of $r_i$'s, you can make $sum_i(q_i-r_i)=infty$ while $sum_i r_i$ converge.
$endgroup$
– Sangchul Lee
Jan 10 at 18:48
$begingroup$
@MathLover nice, thank you
$endgroup$
– nakajuice
Jan 10 at 18:56