Sum of Infinite series with a Geometric series in multiplyThe $n$th derivative of $f(e^x)$Winning Probability / Dice GameFinding formula for the nth partial sumPartial sum of a geometric seriesFormula for partial sum of (10)/(10n+1)Telescoping series order.Proof that a sum is monotonically decreasingSum of infinite series $sum_i=1^infty frac 1 i(i+1)(i+2)…(i+n)$Sequence of Riccati's rootsIs the series $sum_n=1^infty Bigl(1-Bigl(1-frac1n^1+epsilonBigr)^nBigr)$ convergent?Given $S_n = sum dots$ and $a_n = sum dots$ prove that $a_n = S_n + 1over ncdot n!$General formula for a SUM
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Sum of Infinite series with a Geometric series in multiply
The $n$th derivative of $f(e^x)$Winning Probability / Dice GameFinding formula for the nth partial sumPartial sum of a geometric seriesFormula for partial sum of (10)/(10n+1)Telescoping series order.Proof that a sum is monotonically decreasingSum of infinite series $sum_i=1^infty frac 1 i(i+1)(i+2)…(i+n)$Sequence of Riccati's rootsIs the series $sum_n=1^infty Bigl(1-Bigl(1-frac1n^1+epsilonBigr)^nBigr)$ convergent?Given $S_n = sum dots$ and $a_n = sum dots$ prove that $a_n = S_n + 1over ncdot n!$General formula for a SUM
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I came across these series while solving a probability question.
enter link description here
let |r| < 1 ,
$$S_n=sum_k=1^inftyk^n.r^k$$
For n=0 ,it's a GP. $S_0=frac11-r$
For n=1 ,it's a AGP , $S_1=frac-11-r+frac1(1-r)^2$
for n=2 , This series can be reduced to AGP by substituting $k^2=1.3.5....(2k-1)$ & $S_2=frac-r(1-r)^2+frac2r(1-r)^3$.
Is it possible to find sum further in this series . Is there any pattern.
sequences-and-series
asked 9 hours ago
RishiRishi
646 bronze badges
$endgroup$
add a comment |
$begingroup$
I came across these series while solving a probability question.
enter link description here
let |r| < 1 ,
$$S_n=sum_k=1^inftyk^n.r^k$$
For n=0 ,it's a GP. $S_0=frac11-r$
For n=1 ,it's a AGP , $S_1=frac-11-r+frac1(1-r)^2$
for n=2 , This series can be reduced to AGP by substituting $k^2=1.3.5....(2k-1)$ & $S_2=frac-r(1-r)^2+frac2r(1-r)^3$.
Is it possible to find sum further in this series . Is there any pattern.
sequences-and-series
asked 9 hours ago
RishiRishi
646 bronze badges
$endgroup$
add a comment |
$begingroup$
I came across these series while solving a probability question.
enter link description here
let |r| < 1 ,
$$S_n=sum_k=1^inftyk^n.r^k$$
For n=0 ,it's a GP. $S_0=frac11-r$
For n=1 ,it's a AGP , $S_1=frac-11-r+frac1(1-r)^2$
for n=2 , This series can be reduced to AGP by substituting $k^2=1.3.5....(2k-1)$ & $S_2=frac-r(1-r)^2+frac2r(1-r)^3$.
Is it possible to find sum further in this series . Is there any pattern.
sequences-and-series
asked 9 hours ago
RishiRishi
646 bronze badges
$endgroup$
I came across these series while solving a probability question.
enter link description here
let |r| < 1 ,
$$S_n=sum_k=1^inftyk^n.r^k$$
For n=0 ,it's a GP. $S_0=frac11-r$
For n=1 ,it's a AGP , $S_1=frac-11-r+frac1(1-r)^2$
for n=2 , This series can be reduced to AGP by substituting $k^2=1.3.5....(2k-1)$ & $S_2=frac-r(1-r)^2+frac2r(1-r)^3$.
Is it possible to find sum further in this series . Is there any pattern.
sequences-and-series
sequences-and-series
asked 9 hours ago
RishiRishi
646 bronze badges
asked 9 hours ago
RishiRishi
646 bronze badges
edited 8 hours ago
Rishi
asked 9 hours ago
RishiRishi
646 bronze badges
asked 9 hours ago
RishiRishi
646 bronze badges
asked 9 hours ago
RishiRishi
646 bronze badges
646 bronze badges
add a comment |
add a comment |
3 Answers
3
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oldest
votes
$begingroup$
If you take the (formal) dereivative of $S_n$ with respect to $r$, then
$$ fracmathrm dmathrm drS_n=sum_k=1^infty k^nfracmathrm dmathrm drr^k=sum_k=1^infty k^ncdot k r^k-1=frac 1rsum_k=1^ infty k^n+1r^k=frac1rS_n+1$$
so you obtain a recursion formula,
$$S_n+1=rfracmathrm dmathrm drS_n. $$
answered 9 hours ago
Hagen von EitzenHagen von Eitzen
296k24 gold badges285 silver badges522 bronze badges
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$begingroup$
fantastic observation! any connection with riemann zeta?
$endgroup$
– vidyarthi
9 hours ago
1
$begingroup$
@vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
$endgroup$
– Simply Beautiful Art
9 hours ago
add a comment |
$begingroup$
Using the $n$th derivative of $f(e^x)$, we have
$$S_n(e^x)=fracmathrm d^nmathrm dx^nfrac11-e^x=sum_k=0^nnbrace kfrack!e^kx(1-e^x)^k+1$$
and by extension,
$$S_n(r)=sum_k=0^nnbrace kfrack!r^k(1-r)^k+1$$
where $displaystylenbrace k$ are the Stirling numbers of the second kind.
answered 9 hours ago
Simply Beautiful ArtSimply Beautiful Art
53.7k6 gold badges85 silver badges195 bronze badges
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add a comment |
$begingroup$
Given $S_n=sum_k=1^inftyk^ncdot r^k$, it must be:
$$beginalignS_0&=sum_k=1^inftyr^k=fracr1-r=-1+frac11-r; \ S_0'&=frac1(1-r)^2=colorred1(S_0+1)^2;\
S_1&=sum_k=1^inftykcdot r^k=rS_0'=r(S_0+1)^2; \
S_1'&=colorred1(S_0+1)^2+colorred2r(S_0+1)^3\
S_2&=sum_k=1^inftyk^2cdot r^k=rS_1'=r(S_0+1)^2+2r^2(S_0+1)^3; \ S_2'&=colorred1(S_0+1)^2+colorred6r(S_0+1)^3+colorred6r^2(S_0+1)^4\
S_3&=sum_k=1^inftyk^3cdot r^k=rS_2'; \ S_3'&=colorred1(S_0+1)^2+colorred14r(S_0+1)^3+colorred36r^2(S_0+1)^4+colorred24r^3(S_0+1)^5endalign$$
where the coefficients in red are the number sequence A019538.
answered 3 hours ago
farruhotafarruhota
25.2k2 gold badges10 silver badges46 bronze badges
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
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votes
$begingroup$
If you take the (formal) dereivative of $S_n$ with respect to $r$, then
$$ fracmathrm dmathrm drS_n=sum_k=1^infty k^nfracmathrm dmathrm drr^k=sum_k=1^infty k^ncdot k r^k-1=frac 1rsum_k=1^ infty k^n+1r^k=frac1rS_n+1$$
so you obtain a recursion formula,
$$S_n+1=rfracmathrm dmathrm drS_n. $$
answered 9 hours ago
Hagen von EitzenHagen von Eitzen
296k24 gold badges285 silver badges522 bronze badges
$endgroup$
$begingroup$
fantastic observation! any connection with riemann zeta?
$endgroup$
– vidyarthi
9 hours ago
1
$begingroup$
@vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
$endgroup$
– Simply Beautiful Art
9 hours ago
add a comment |
$begingroup$
If you take the (formal) dereivative of $S_n$ with respect to $r$, then
$$ fracmathrm dmathrm drS_n=sum_k=1^infty k^nfracmathrm dmathrm drr^k=sum_k=1^infty k^ncdot k r^k-1=frac 1rsum_k=1^ infty k^n+1r^k=frac1rS_n+1$$
so you obtain a recursion formula,
$$S_n+1=rfracmathrm dmathrm drS_n. $$
answered 9 hours ago
Hagen von EitzenHagen von Eitzen
296k24 gold badges285 silver badges522 bronze badges
$endgroup$
$begingroup$
fantastic observation! any connection with riemann zeta?
$endgroup$
– vidyarthi
9 hours ago
1
$begingroup$
@vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
$endgroup$
– Simply Beautiful Art
9 hours ago
add a comment |
$begingroup$
If you take the (formal) dereivative of $S_n$ with respect to $r$, then
$$ fracmathrm dmathrm drS_n=sum_k=1^infty k^nfracmathrm dmathrm drr^k=sum_k=1^infty k^ncdot k r^k-1=frac 1rsum_k=1^ infty k^n+1r^k=frac1rS_n+1$$
so you obtain a recursion formula,
$$S_n+1=rfracmathrm dmathrm drS_n. $$
answered 9 hours ago
Hagen von EitzenHagen von Eitzen
296k24 gold badges285 silver badges522 bronze badges
$endgroup$
If you take the (formal) dereivative of $S_n$ with respect to $r$, then
$$ fracmathrm dmathrm drS_n=sum_k=1^infty k^nfracmathrm dmathrm drr^k=sum_k=1^infty k^ncdot k r^k-1=frac 1rsum_k=1^ infty k^n+1r^k=frac1rS_n+1$$
so you obtain a recursion formula,
$$S_n+1=rfracmathrm dmathrm drS_n. $$
answered 9 hours ago
Hagen von EitzenHagen von Eitzen
296k24 gold badges285 silver badges522 bronze badges
answered 9 hours ago
Hagen von EitzenHagen von Eitzen
296k24 gold badges285 silver badges522 bronze badges
answered 9 hours ago
Hagen von EitzenHagen von Eitzen
296k24 gold badges285 silver badges522 bronze badges
answered 9 hours ago
Hagen von EitzenHagen von Eitzen
296k24 gold badges285 silver badges522 bronze badges
296k24 gold badges285 silver badges522 bronze badges
$begingroup$
fantastic observation! any connection with riemann zeta?
$endgroup$
– vidyarthi
9 hours ago
1
$begingroup$
@vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
$endgroup$
– Simply Beautiful Art
9 hours ago
add a comment |
$begingroup$
fantastic observation! any connection with riemann zeta?
$endgroup$
– vidyarthi
9 hours ago
1
$begingroup$
@vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
$endgroup$
– Simply Beautiful Art
9 hours ago
$begingroup$
fantastic observation! any connection with riemann zeta?
$endgroup$
– vidyarthi
9 hours ago
$begingroup$
fantastic observation! any connection with riemann zeta?
$endgroup$
– vidyarthi
9 hours ago
1
1
$begingroup$
@vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
$endgroup$
– Simply Beautiful Art
9 hours ago
$begingroup$
@vidyarthi You can compute the Dirichlet eta function, and hence the Riemann zeta function, by taking the limit as $rto-1$.
$endgroup$
– Simply Beautiful Art
9 hours ago
add a comment |
$begingroup$
Using the $n$th derivative of $f(e^x)$, we have
$$S_n(e^x)=fracmathrm d^nmathrm dx^nfrac11-e^x=sum_k=0^nnbrace kfrack!e^kx(1-e^x)^k+1$$
and by extension,
$$S_n(r)=sum_k=0^nnbrace kfrack!r^k(1-r)^k+1$$
where $displaystylenbrace k$ are the Stirling numbers of the second kind.
answered 9 hours ago
Simply Beautiful ArtSimply Beautiful Art
53.7k6 gold badges85 silver badges195 bronze badges
$endgroup$
add a comment |
$begingroup$
Using the $n$th derivative of $f(e^x)$, we have
$$S_n(e^x)=fracmathrm d^nmathrm dx^nfrac11-e^x=sum_k=0^nnbrace kfrack!e^kx(1-e^x)^k+1$$
and by extension,
$$S_n(r)=sum_k=0^nnbrace kfrack!r^k(1-r)^k+1$$
where $displaystylenbrace k$ are the Stirling numbers of the second kind.
answered 9 hours ago
Simply Beautiful ArtSimply Beautiful Art
53.7k6 gold badges85 silver badges195 bronze badges
$endgroup$
add a comment |
$begingroup$
Using the $n$th derivative of $f(e^x)$, we have
$$S_n(e^x)=fracmathrm d^nmathrm dx^nfrac11-e^x=sum_k=0^nnbrace kfrack!e^kx(1-e^x)^k+1$$
and by extension,
$$S_n(r)=sum_k=0^nnbrace kfrack!r^k(1-r)^k+1$$
where $displaystylenbrace k$ are the Stirling numbers of the second kind.
answered 9 hours ago
Simply Beautiful ArtSimply Beautiful Art
53.7k6 gold badges85 silver badges195 bronze badges
$endgroup$
Using the $n$th derivative of $f(e^x)$, we have
$$S_n(e^x)=fracmathrm d^nmathrm dx^nfrac11-e^x=sum_k=0^nnbrace kfrack!e^kx(1-e^x)^k+1$$
and by extension,
$$S_n(r)=sum_k=0^nnbrace kfrack!r^k(1-r)^k+1$$
where $displaystylenbrace k$ are the Stirling numbers of the second kind.
answered 9 hours ago
Simply Beautiful ArtSimply Beautiful Art
53.7k6 gold badges85 silver badges195 bronze badges
answered 9 hours ago
Simply Beautiful ArtSimply Beautiful Art
53.7k6 gold badges85 silver badges195 bronze badges
answered 9 hours ago
Simply Beautiful ArtSimply Beautiful Art
53.7k6 gold badges85 silver badges195 bronze badges
answered 9 hours ago
Simply Beautiful ArtSimply Beautiful Art
53.7k6 gold badges85 silver badges195 bronze badges
53.7k6 gold badges85 silver badges195 bronze badges
add a comment |
add a comment |
$begingroup$
Given $S_n=sum_k=1^inftyk^ncdot r^k$, it must be:
$$beginalignS_0&=sum_k=1^inftyr^k=fracr1-r=-1+frac11-r; \ S_0'&=frac1(1-r)^2=colorred1(S_0+1)^2;\
S_1&=sum_k=1^inftykcdot r^k=rS_0'=r(S_0+1)^2; \
S_1'&=colorred1(S_0+1)^2+colorred2r(S_0+1)^3\
S_2&=sum_k=1^inftyk^2cdot r^k=rS_1'=r(S_0+1)^2+2r^2(S_0+1)^3; \ S_2'&=colorred1(S_0+1)^2+colorred6r(S_0+1)^3+colorred6r^2(S_0+1)^4\
S_3&=sum_k=1^inftyk^3cdot r^k=rS_2'; \ S_3'&=colorred1(S_0+1)^2+colorred14r(S_0+1)^3+colorred36r^2(S_0+1)^4+colorred24r^3(S_0+1)^5endalign$$
where the coefficients in red are the number sequence A019538.
answered 3 hours ago
farruhotafarruhota
25.2k2 gold badges10 silver badges46 bronze badges
$endgroup$
add a comment |
$begingroup$
Given $S_n=sum_k=1^inftyk^ncdot r^k$, it must be:
$$beginalignS_0&=sum_k=1^inftyr^k=fracr1-r=-1+frac11-r; \ S_0'&=frac1(1-r)^2=colorred1(S_0+1)^2;\
S_1&=sum_k=1^inftykcdot r^k=rS_0'=r(S_0+1)^2; \
S_1'&=colorred1(S_0+1)^2+colorred2r(S_0+1)^3\
S_2&=sum_k=1^inftyk^2cdot r^k=rS_1'=r(S_0+1)^2+2r^2(S_0+1)^3; \ S_2'&=colorred1(S_0+1)^2+colorred6r(S_0+1)^3+colorred6r^2(S_0+1)^4\
S_3&=sum_k=1^inftyk^3cdot r^k=rS_2'; \ S_3'&=colorred1(S_0+1)^2+colorred14r(S_0+1)^3+colorred36r^2(S_0+1)^4+colorred24r^3(S_0+1)^5endalign$$
where the coefficients in red are the number sequence A019538.
answered 3 hours ago
farruhotafarruhota
25.2k2 gold badges10 silver badges46 bronze badges
$endgroup$
add a comment |
$begingroup$
Given $S_n=sum_k=1^inftyk^ncdot r^k$, it must be:
$$beginalignS_0&=sum_k=1^inftyr^k=fracr1-r=-1+frac11-r; \ S_0'&=frac1(1-r)^2=colorred1(S_0+1)^2;\
S_1&=sum_k=1^inftykcdot r^k=rS_0'=r(S_0+1)^2; \
S_1'&=colorred1(S_0+1)^2+colorred2r(S_0+1)^3\
S_2&=sum_k=1^inftyk^2cdot r^k=rS_1'=r(S_0+1)^2+2r^2(S_0+1)^3; \ S_2'&=colorred1(S_0+1)^2+colorred6r(S_0+1)^3+colorred6r^2(S_0+1)^4\
S_3&=sum_k=1^inftyk^3cdot r^k=rS_2'; \ S_3'&=colorred1(S_0+1)^2+colorred14r(S_0+1)^3+colorred36r^2(S_0+1)^4+colorred24r^3(S_0+1)^5endalign$$
where the coefficients in red are the number sequence A019538.
answered 3 hours ago
farruhotafarruhota
25.2k2 gold badges10 silver badges46 bronze badges
$endgroup$
Given $S_n=sum_k=1^inftyk^ncdot r^k$, it must be:
$$beginalignS_0&=sum_k=1^inftyr^k=fracr1-r=-1+frac11-r; \ S_0'&=frac1(1-r)^2=colorred1(S_0+1)^2;\
S_1&=sum_k=1^inftykcdot r^k=rS_0'=r(S_0+1)^2; \
S_1'&=colorred1(S_0+1)^2+colorred2r(S_0+1)^3\
S_2&=sum_k=1^inftyk^2cdot r^k=rS_1'=r(S_0+1)^2+2r^2(S_0+1)^3; \ S_2'&=colorred1(S_0+1)^2+colorred6r(S_0+1)^3+colorred6r^2(S_0+1)^4\
S_3&=sum_k=1^inftyk^3cdot r^k=rS_2'; \ S_3'&=colorred1(S_0+1)^2+colorred14r(S_0+1)^3+colorred36r^2(S_0+1)^4+colorred24r^3(S_0+1)^5endalign$$
where the coefficients in red are the number sequence A019538.
answered 3 hours ago
farruhotafarruhota
25.2k2 gold badges10 silver badges46 bronze badges
answered 3 hours ago
farruhotafarruhota
25.2k2 gold badges10 silver badges46 bronze badges
answered 3 hours ago
farruhotafarruhota
25.2k2 gold badges10 silver badges46 bronze badges
answered 3 hours ago
farruhotafarruhota
25.2k2 gold badges10 silver badges46 bronze badges
25.2k2 gold badges10 silver badges46 bronze badges
add a comment |
add a comment |
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