Does this sum go infinity? [on hold]If $a_n$ goes to zero, can we find signs $s_n$ such that $sum s_n a_n$ converges?Asymptotic expansion for harmonic sum in two variablesConvergence of $sum_k=1^n(1-k/n)a_k$Does $sum_n=1^inftyfraccosleft(fracnpi2right)sqrtn$ converge?Bivariate infinite series: explicit sum?some infinite sum and $liminf$Sum of the inverses of numbers with $n$ divisors.Why $sum_n=0^infty (-1)^nx^2n$ converge pointwise?Does this series converge absolutely $sum_n=1^inftyfracb^n_ncos(npi)n$Does $sumlimits_k=1^infty sumlimits_n=k^infty frac(-1)^n+kn$ diverge?
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Does this sum go infinity? [on hold]
If $a_n$ goes to zero, can we find signs $s_n$ such that $sum s_n a_n$ converges?Asymptotic expansion for harmonic sum in two variablesConvergence of $sum_k=1^n(1-k/n)a_k$Does $sum_n=1^inftyfraccosleft(fracnpi2right)sqrtn$ converge?Bivariate infinite series: explicit sum?some infinite sum and $liminf$Sum of the inverses of numbers with $n$ divisors.Why $sum_n=0^infty (-1)^nx^2n$ converge pointwise?Does this series converge absolutely $sum_n=1^inftyfracb^n_ncos(npi)n$Does $sumlimits_k=1^infty sumlimits_n=k^infty frac(-1)^n+kn$ diverge?
$begingroup$
Consider $F(x)$ that maps from $Bbb N$ to $pm 1$, such that if $x$ is odd, then $F(x)$ = $$(-1)^(fracx-12)$$, and if $x$ is even, then $F(x)=F(y)$, where $y$ is the odd number obtained after dividing $x$ by $2$ until it is odd.
Does $S_n = sum_p=1^n F(p)$ have an explicit formula?
And if $n$ tends to $infty$ does the sum alternates or will it lie between an interval or does it tend to $pm infty$ ?
sequences-and-series algebra-precalculus
asked 2 days ago
Hari Krishna PHari Krishna P
385
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Hari Krishna P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel
add a comment |
$begingroup$
Consider $F(x)$ that maps from $Bbb N$ to $pm 1$, such that if $x$ is odd, then $F(x)$ = $$(-1)^(fracx-12)$$, and if $x$ is even, then $F(x)=F(y)$, where $y$ is the odd number obtained after dividing $x$ by $2$ until it is odd.
Does $S_n = sum_p=1^n F(p)$ have an explicit formula?
And if $n$ tends to $infty$ does the sum alternates or will it lie between an interval or does it tend to $pm infty$ ?
sequences-and-series algebra-precalculus
asked 2 days ago
Hari Krishna PHari Krishna P
385
New contributor
Hari Krishna P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel
1
$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
2 days ago
add a comment |
$begingroup$
Consider $F(x)$ that maps from $Bbb N$ to $pm 1$, such that if $x$ is odd, then $F(x)$ = $$(-1)^(fracx-12)$$, and if $x$ is even, then $F(x)=F(y)$, where $y$ is the odd number obtained after dividing $x$ by $2$ until it is odd.
Does $S_n = sum_p=1^n F(p)$ have an explicit formula?
And if $n$ tends to $infty$ does the sum alternates or will it lie between an interval or does it tend to $pm infty$ ?
sequences-and-series algebra-precalculus
asked 2 days ago
Hari Krishna PHari Krishna P
385
New contributor
Hari Krishna P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Consider $F(x)$ that maps from $Bbb N$ to $pm 1$, such that if $x$ is odd, then $F(x)$ = $$(-1)^(fracx-12)$$, and if $x$ is even, then $F(x)=F(y)$, where $y$ is the odd number obtained after dividing $x$ by $2$ until it is odd.
Does $S_n = sum_p=1^n F(p)$ have an explicit formula?
And if $n$ tends to $infty$ does the sum alternates or will it lie between an interval or does it tend to $pm infty$ ?
sequences-and-series algebra-precalculus
sequences-and-series algebra-precalculus
asked 2 days ago
Hari Krishna PHari Krishna P
385
New contributor
Hari Krishna P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Hari Krishna PHari Krishna P
385
New contributor
Hari Krishna P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Hari Krishna P
asked 2 days ago
Hari Krishna PHari Krishna P
385
New contributor
Hari Krishna P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Hari Krishna PHari Krishna P
385
asked 2 days ago
Hari Krishna PHari Krishna P
385
385
New contributor
Hari Krishna P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Hari Krishna P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hari Krishna P is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel
put on hold as off-topic by RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel
1
$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
2 days ago
add a comment |
1
$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
2 days ago
1
1
$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
2 days ago
$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $T(n) = sum_ktext is odd, kle nF(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.
Then, $T(lfloor n/2rfloor) = sum_ktext is odd, kle lfloor n/2rfloorF(k)=sum_ktext is odd, 2kle nF(2k)$, $T(lfloor n/4rfloor) = sum_ktext is odd, 4kle nF(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_m=0^lfloor log_2 nrfloorT(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.
Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^m-1$ place, is always there) or as high as $m$ for $n=lfloor 2^m+1/3rfloor$.
So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.
answered 2 days ago
jmerryjmerry
14.8k1632
$endgroup$
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $T(n) = sum_ktext is odd, kle nF(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.
Then, $T(lfloor n/2rfloor) = sum_ktext is odd, kle lfloor n/2rfloorF(k)=sum_ktext is odd, 2kle nF(2k)$, $T(lfloor n/4rfloor) = sum_ktext is odd, 4kle nF(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_m=0^lfloor log_2 nrfloorT(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.
Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^m-1$ place, is always there) or as high as $m$ for $n=lfloor 2^m+1/3rfloor$.
So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.
answered 2 days ago
jmerryjmerry
14.8k1632
$endgroup$
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
2 days ago
add a comment |
$begingroup$
Let $T(n) = sum_ktext is odd, kle nF(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.
Then, $T(lfloor n/2rfloor) = sum_ktext is odd, kle lfloor n/2rfloorF(k)=sum_ktext is odd, 2kle nF(2k)$, $T(lfloor n/4rfloor) = sum_ktext is odd, 4kle nF(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_m=0^lfloor log_2 nrfloorT(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.
Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^m-1$ place, is always there) or as high as $m$ for $n=lfloor 2^m+1/3rfloor$.
So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.
answered 2 days ago
jmerryjmerry
14.8k1632
$endgroup$
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
2 days ago
add a comment |
$begingroup$
Let $T(n) = sum_ktext is odd, kle nF(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.
Then, $T(lfloor n/2rfloor) = sum_ktext is odd, kle lfloor n/2rfloorF(k)=sum_ktext is odd, 2kle nF(2k)$, $T(lfloor n/4rfloor) = sum_ktext is odd, 4kle nF(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_m=0^lfloor log_2 nrfloorT(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.
Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^m-1$ place, is always there) or as high as $m$ for $n=lfloor 2^m+1/3rfloor$.
So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.
answered 2 days ago
jmerryjmerry
14.8k1632
$endgroup$
Let $T(n) = sum_ktext is odd, kle nF(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.
Then, $T(lfloor n/2rfloor) = sum_ktext is odd, kle lfloor n/2rfloorF(k)=sum_ktext is odd, 2kle nF(2k)$, $T(lfloor n/4rfloor) = sum_ktext is odd, 4kle nF(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_m=0^lfloor log_2 nrfloorT(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.
Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^m-1$ place, is always there) or as high as $m$ for $n=lfloor 2^m+1/3rfloor$.
So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.
answered 2 days ago
jmerryjmerry
14.8k1632
answered 2 days ago
jmerryjmerry
14.8k1632
answered 2 days ago
jmerryjmerry
14.8k1632
answered 2 days ago
jmerryjmerry
14.8k1632
14.8k1632
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
2 days ago
add a comment |
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
2 days ago
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
2 days ago
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
2 days ago
add a comment |
1
$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
2 days ago