Is sum of the reciprocal of any subsequence of natural number divergent?
$begingroup$
Let, ${a_n}$ be any subsequence of natural numbers.
Then, is $sumfrac{1}{a_n}$ always divergent?
sequences-and-series
asked Dec 6 '18 at 5:31
Tom.Tom.
14218
$endgroup$
add a comment |
$begingroup$
Let, ${a_n}$ be any subsequence of natural numbers.
Then, is $sumfrac{1}{a_n}$ always divergent?
sequences-and-series
asked Dec 6 '18 at 5:31
Tom.Tom.
14218
$endgroup$
2
$begingroup$
This is almost like asking if there are any series that converge...
$endgroup$
– Clayton
Dec 6 '18 at 5:33
add a comment |
$begingroup$
Let, ${a_n}$ be any subsequence of natural numbers.
Then, is $sumfrac{1}{a_n}$ always divergent?
sequences-and-series
asked Dec 6 '18 at 5:31
Tom.Tom.
14218
$endgroup$
Let, ${a_n}$ be any subsequence of natural numbers.
Then, is $sumfrac{1}{a_n}$ always divergent?
sequences-and-series
sequences-and-series
asked Dec 6 '18 at 5:31
Tom.Tom.
14218
asked Dec 6 '18 at 5:31
Tom.Tom.
14218
edited Dec 6 '18 at 5:34
Tom.
asked Dec 6 '18 at 5:31
Tom.Tom.
14218
asked Dec 6 '18 at 5:31
Tom.Tom.
14218
asked Dec 6 '18 at 5:31
Tom.Tom.
14218
14218
2
$begingroup$
This is almost like asking if there are any series that converge...
$endgroup$
– Clayton
Dec 6 '18 at 5:33
add a comment |
2
$begingroup$
This is almost like asking if there are any series that converge...
$endgroup$
– Clayton
Dec 6 '18 at 5:33
2
2
$begingroup$
This is almost like asking if there are any series that converge...
$endgroup$
– Clayton
Dec 6 '18 at 5:33
$begingroup$
This is almost like asking if there are any series that converge...
$endgroup$
– Clayton
Dec 6 '18 at 5:33
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
No. Consider the subsequence defined by the powers of two: ${2,4,8,16,32,64,...}$. This results in the summation
$$sum_{k=1}^infty frac{1}{2^k} = frac{1}{2} +frac{1}{4} +frac{1}{8} +frac{1}{16} +frac{1}{32} + ...$$
We note that this is an infinite geometric series of ratio $1/2$. Thus, from the formula for the summation of such a series, it can be shown that this summation converges to $1$.
Truly many such series exist; this is just one such example.
answered Dec 6 '18 at 5:34
Eevee TrainerEevee Trainer
5,4491936
$endgroup$
add a comment |
$begingroup$
No. It is easy to produce a counter example
Let $a_n=n^2$. So it is the sequence of square numbers 1, 4, 9 ,16 ,..
Then it is known that:
$$sum_{n=1}^infty frac 1{n^2} = frac{pi^2}{6}$$
answered Dec 6 '18 at 5:35
M.AM.A
1599
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add a comment |
$begingroup$
$sum_{k=1}^{infty}frac{1}{k^p}:left{begin{matrix}
mathrm{converges} &,p>1 \
mathrm{diverges} &,pleq1
end{matrix}right.$
answered Dec 6 '18 at 5:47
Hussain-AlqatariHussain-Alqatari
3187
$endgroup$
add a comment |
$begingroup$
No. Even though $sum 1/n$ diverges,
if you discard all the terms whose denominator involves the digit $9$, the series would become convergent, and the sum would be less than $80$.
answered Dec 6 '18 at 5:47
xbhxbh
6,0231522
$endgroup$
$begingroup$
Several correct answers to this question, but this is the only cute answer (so far).
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 10:15
add a comment |
$begingroup$
Let $n$ be any natural number greater than $1$,
$$sum_{k=0}^inftyfrac{1}{n^k}=frac{1}{1-frac1n}=frac{n}{n-1}$$
is a geometric series that converges.
answered Dec 6 '18 at 7:49
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. Consider the subsequence defined by the powers of two: ${2,4,8,16,32,64,...}$. This results in the summation
$$sum_{k=1}^infty frac{1}{2^k} = frac{1}{2} +frac{1}{4} +frac{1}{8} +frac{1}{16} +frac{1}{32} + ...$$
We note that this is an infinite geometric series of ratio $1/2$. Thus, from the formula for the summation of such a series, it can be shown that this summation converges to $1$.
Truly many such series exist; this is just one such example.
answered Dec 6 '18 at 5:34
Eevee TrainerEevee Trainer
5,4491936
$endgroup$
add a comment |
$begingroup$
No. Consider the subsequence defined by the powers of two: ${2,4,8,16,32,64,...}$. This results in the summation
$$sum_{k=1}^infty frac{1}{2^k} = frac{1}{2} +frac{1}{4} +frac{1}{8} +frac{1}{16} +frac{1}{32} + ...$$
We note that this is an infinite geometric series of ratio $1/2$. Thus, from the formula for the summation of such a series, it can be shown that this summation converges to $1$.
Truly many such series exist; this is just one such example.
answered Dec 6 '18 at 5:34
Eevee TrainerEevee Trainer
5,4491936
$endgroup$
add a comment |
$begingroup$
No. Consider the subsequence defined by the powers of two: ${2,4,8,16,32,64,...}$. This results in the summation
$$sum_{k=1}^infty frac{1}{2^k} = frac{1}{2} +frac{1}{4} +frac{1}{8} +frac{1}{16} +frac{1}{32} + ...$$
We note that this is an infinite geometric series of ratio $1/2$. Thus, from the formula for the summation of such a series, it can be shown that this summation converges to $1$.
Truly many such series exist; this is just one such example.
answered Dec 6 '18 at 5:34
Eevee TrainerEevee Trainer
5,4491936
$endgroup$
No. Consider the subsequence defined by the powers of two: ${2,4,8,16,32,64,...}$. This results in the summation
$$sum_{k=1}^infty frac{1}{2^k} = frac{1}{2} +frac{1}{4} +frac{1}{8} +frac{1}{16} +frac{1}{32} + ...$$
We note that this is an infinite geometric series of ratio $1/2$. Thus, from the formula for the summation of such a series, it can be shown that this summation converges to $1$.
Truly many such series exist; this is just one such example.
answered Dec 6 '18 at 5:34
Eevee TrainerEevee Trainer
5,4491936
answered Dec 6 '18 at 5:34
Eevee TrainerEevee Trainer
5,4491936
answered Dec 6 '18 at 5:34
Eevee TrainerEevee Trainer
5,4491936
answered Dec 6 '18 at 5:34
Eevee TrainerEevee Trainer
5,4491936
5,4491936
add a comment |
add a comment |
$begingroup$
No. It is easy to produce a counter example
Let $a_n=n^2$. So it is the sequence of square numbers 1, 4, 9 ,16 ,..
Then it is known that:
$$sum_{n=1}^infty frac 1{n^2} = frac{pi^2}{6}$$
answered Dec 6 '18 at 5:35
M.AM.A
1599
$endgroup$
add a comment |
$begingroup$
No. It is easy to produce a counter example
Let $a_n=n^2$. So it is the sequence of square numbers 1, 4, 9 ,16 ,..
Then it is known that:
$$sum_{n=1}^infty frac 1{n^2} = frac{pi^2}{6}$$
answered Dec 6 '18 at 5:35
M.AM.A
1599
$endgroup$
add a comment |
$begingroup$
No. It is easy to produce a counter example
Let $a_n=n^2$. So it is the sequence of square numbers 1, 4, 9 ,16 ,..
Then it is known that:
$$sum_{n=1}^infty frac 1{n^2} = frac{pi^2}{6}$$
answered Dec 6 '18 at 5:35
M.AM.A
1599
$endgroup$
No. It is easy to produce a counter example
Let $a_n=n^2$. So it is the sequence of square numbers 1, 4, 9 ,16 ,..
Then it is known that:
$$sum_{n=1}^infty frac 1{n^2} = frac{pi^2}{6}$$
answered Dec 6 '18 at 5:35
M.AM.A
1599
answered Dec 6 '18 at 5:35
M.AM.A
1599
answered Dec 6 '18 at 5:35
M.AM.A
1599
answered Dec 6 '18 at 5:35
M.AM.A
1599
1599
add a comment |
add a comment |
$begingroup$
$sum_{k=1}^{infty}frac{1}{k^p}:left{begin{matrix}
mathrm{converges} &,p>1 \
mathrm{diverges} &,pleq1
end{matrix}right.$
answered Dec 6 '18 at 5:47
Hussain-AlqatariHussain-Alqatari
3187
$endgroup$
add a comment |
$begingroup$
$sum_{k=1}^{infty}frac{1}{k^p}:left{begin{matrix}
mathrm{converges} &,p>1 \
mathrm{diverges} &,pleq1
end{matrix}right.$
answered Dec 6 '18 at 5:47
Hussain-AlqatariHussain-Alqatari
3187
$endgroup$
add a comment |
$begingroup$
$sum_{k=1}^{infty}frac{1}{k^p}:left{begin{matrix}
mathrm{converges} &,p>1 \
mathrm{diverges} &,pleq1
end{matrix}right.$
answered Dec 6 '18 at 5:47
Hussain-AlqatariHussain-Alqatari
3187
$endgroup$
$sum_{k=1}^{infty}frac{1}{k^p}:left{begin{matrix}
mathrm{converges} &,p>1 \
mathrm{diverges} &,pleq1
end{matrix}right.$
answered Dec 6 '18 at 5:47
Hussain-AlqatariHussain-Alqatari
3187
answered Dec 6 '18 at 5:47
Hussain-AlqatariHussain-Alqatari
3187
answered Dec 6 '18 at 5:47
Hussain-AlqatariHussain-Alqatari
3187
answered Dec 6 '18 at 5:47
Hussain-AlqatariHussain-Alqatari
3187
3187
add a comment |
add a comment |
$begingroup$
No. Even though $sum 1/n$ diverges,
if you discard all the terms whose denominator involves the digit $9$, the series would become convergent, and the sum would be less than $80$.
answered Dec 6 '18 at 5:47
xbhxbh
6,0231522
$endgroup$
$begingroup$
Several correct answers to this question, but this is the only cute answer (so far).
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 10:15
add a comment |
$begingroup$
No. Even though $sum 1/n$ diverges,
if you discard all the terms whose denominator involves the digit $9$, the series would become convergent, and the sum would be less than $80$.
answered Dec 6 '18 at 5:47
xbhxbh
6,0231522
$endgroup$
$begingroup$
Several correct answers to this question, but this is the only cute answer (so far).
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 10:15
add a comment |
$begingroup$
No. Even though $sum 1/n$ diverges,
if you discard all the terms whose denominator involves the digit $9$, the series would become convergent, and the sum would be less than $80$.
answered Dec 6 '18 at 5:47
xbhxbh
6,0231522
$endgroup$
No. Even though $sum 1/n$ diverges,
if you discard all the terms whose denominator involves the digit $9$, the series would become convergent, and the sum would be less than $80$.
answered Dec 6 '18 at 5:47
xbhxbh
6,0231522
answered Dec 6 '18 at 5:47
xbhxbh
6,0231522
answered Dec 6 '18 at 5:47
xbhxbh
6,0231522
answered Dec 6 '18 at 5:47
xbhxbh
6,0231522
6,0231522
$begingroup$
Several correct answers to this question, but this is the only cute answer (so far).
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 10:15
add a comment |
$begingroup$
Several correct answers to this question, but this is the only cute answer (so far).
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 10:15
$begingroup$
Several correct answers to this question, but this is the only cute answer (so far).
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 10:15
$begingroup$
Several correct answers to this question, but this is the only cute answer (so far).
$endgroup$
– Jyrki Lahtonen
Dec 6 '18 at 10:15
add a comment |
$begingroup$
Let $n$ be any natural number greater than $1$,
$$sum_{k=0}^inftyfrac{1}{n^k}=frac{1}{1-frac1n}=frac{n}{n-1}$$
is a geometric series that converges.
answered Dec 6 '18 at 7:49
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
Let $n$ be any natural number greater than $1$,
$$sum_{k=0}^inftyfrac{1}{n^k}=frac{1}{1-frac1n}=frac{n}{n-1}$$
is a geometric series that converges.
answered Dec 6 '18 at 7:49
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
Let $n$ be any natural number greater than $1$,
$$sum_{k=0}^inftyfrac{1}{n^k}=frac{1}{1-frac1n}=frac{n}{n-1}$$
is a geometric series that converges.
answered Dec 6 '18 at 7:49
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
Let $n$ be any natural number greater than $1$,
$$sum_{k=0}^inftyfrac{1}{n^k}=frac{1}{1-frac1n}=frac{n}{n-1}$$
is a geometric series that converges.
answered Dec 6 '18 at 7:49
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 6 '18 at 7:49
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 6 '18 at 7:49
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 6 '18 at 7:49
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
add a comment |
add a comment |
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2
$begingroup$
This is almost like asking if there are any series that converge...
$endgroup$
– Clayton
Dec 6 '18 at 5:33