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 12 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 12 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 12 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 12 hours ago
RishiRishi
646 bronze badges
asked 12 hours ago
RishiRishi
646 bronze badges
edited 11 hours ago
Rishi
asked 12 hours ago
RishiRishi
646 bronze badges
asked 12 hours ago
RishiRishi
646 bronze badges
asked 12 hours ago
RishiRishi
646 bronze badges
646 bronze badges
add a comment |
add a comment |
3 Answers
3
active
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 12 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
12 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
11 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 11 hours ago
Simply Beautiful ArtSimply Beautiful Art
53.7k6 gold badges85 silver badges195 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 6 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
oldest
votes
active
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 12 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
12 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
11 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 12 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
12 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
11 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 12 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 12 hours ago
Hagen von EitzenHagen von Eitzen
296k24 gold badges285 silver badges522 bronze badges
answered 12 hours ago
Hagen von EitzenHagen von Eitzen
296k24 gold badges285 silver badges522 bronze badges
answered 12 hours ago
Hagen von EitzenHagen von Eitzen
296k24 gold badges285 silver badges522 bronze badges
answered 12 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
12 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
11 hours ago
add a comment |
$begingroup$
fantastic observation! any connection with riemann zeta?
$endgroup$
– vidyarthi
12 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
11 hours ago
$begingroup$
fantastic observation! any connection with riemann zeta?
$endgroup$
– vidyarthi
12 hours ago
$begingroup$
fantastic observation! any connection with riemann zeta?
$endgroup$
– vidyarthi
12 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
11 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
11 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 11 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 11 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 11 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 11 hours ago
Simply Beautiful ArtSimply Beautiful Art
53.7k6 gold badges85 silver badges195 bronze badges
answered 11 hours ago
Simply Beautiful ArtSimply Beautiful Art
53.7k6 gold badges85 silver badges195 bronze badges
answered 11 hours ago
Simply Beautiful ArtSimply Beautiful Art
53.7k6 gold badges85 silver badges195 bronze badges
answered 11 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 6 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 6 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 6 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 6 hours ago
farruhotafarruhota
25.2k2 gold badges10 silver badges46 bronze badges
answered 6 hours ago
farruhotafarruhota
25.2k2 gold badges10 silver badges46 bronze badges
answered 6 hours ago
farruhotafarruhota
25.2k2 gold badges10 silver badges46 bronze badges
answered 6 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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