Mixing
How to find the value this sum converges to?$sum_{n=2}^{infty}frac{n+1}{n(2n-1)(2n+1)} $
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How to find the value this sum converges to?$$sum_{n=2}^{infty}frac{n+1}{n(2n-1)(2n+1)} $$
I've tried separating it like
$$sum_{n=2}^{infty}frac{1/2}{(2n+1)}+frac{3/2}{(2n-1)}-frac{1}{n}$$
and writing some terms and I get
$$1/6·(1+1/2+...+1/6n+3)+1/2·(1/3+1+1/2+...+1/6n-3)-(1+1/2+...1/n)$$
but I don't know how to end summing it all! Any hint!
FYI I haven't learnt integration and differentation.
real-analysis summation
asked Jan 24 at 22:59
iggykimiiggykimi
30710
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add a comment |
$begingroup$
How to find the value this sum converges to?$$sum_{n=2}^{infty}frac{n+1}{n(2n-1)(2n+1)} $$
I've tried separating it like
$$sum_{n=2}^{infty}frac{1/2}{(2n+1)}+frac{3/2}{(2n-1)}-frac{1}{n}$$
and writing some terms and I get
$$1/6·(1+1/2+...+1/6n+3)+1/2·(1/3+1+1/2+...+1/6n-3)-(1+1/2+...1/n)$$
but I don't know how to end summing it all! Any hint!
FYI I haven't learnt integration and differentation.
real-analysis summation
asked Jan 24 at 22:59
iggykimiiggykimi
30710
$endgroup$
$begingroup$
You divided your convergent series into three divergent ones. Doesn't look very promising...unless you can produce a telescopic series of something like that (hint)
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– DonAntonio
Jan 24 at 23:01
add a comment |
$begingroup$
How to find the value this sum converges to?$$sum_{n=2}^{infty}frac{n+1}{n(2n-1)(2n+1)} $$
I've tried separating it like
$$sum_{n=2}^{infty}frac{1/2}{(2n+1)}+frac{3/2}{(2n-1)}-frac{1}{n}$$
and writing some terms and I get
$$1/6·(1+1/2+...+1/6n+3)+1/2·(1/3+1+1/2+...+1/6n-3)-(1+1/2+...1/n)$$
but I don't know how to end summing it all! Any hint!
FYI I haven't learnt integration and differentation.
real-analysis summation
asked Jan 24 at 22:59
iggykimiiggykimi
30710
$endgroup$
How to find the value this sum converges to?$$sum_{n=2}^{infty}frac{n+1}{n(2n-1)(2n+1)} $$
I've tried separating it like
$$sum_{n=2}^{infty}frac{1/2}{(2n+1)}+frac{3/2}{(2n-1)}-frac{1}{n}$$
and writing some terms and I get
$$1/6·(1+1/2+...+1/6n+3)+1/2·(1/3+1+1/2+...+1/6n-3)-(1+1/2+...1/n)$$
but I don't know how to end summing it all! Any hint!
FYI I haven't learnt integration and differentation.
real-analysis summation
real-analysis summation
asked Jan 24 at 22:59
iggykimiiggykimi
30710
asked Jan 24 at 22:59
iggykimiiggykimi
30710
asked Jan 24 at 22:59
iggykimiiggykimi
30710
asked Jan 24 at 22:59
iggykimiiggykimi
30710
asked Jan 24 at 22:59
iggykimiiggykimi
30710
30710
$begingroup$
You divided your convergent series into three divergent ones. Doesn't look very promising...unless you can produce a telescopic series of something like that (hint)
$endgroup$
– DonAntonio
Jan 24 at 23:01
add a comment |
$begingroup$
You divided your convergent series into three divergent ones. Doesn't look very promising...unless you can produce a telescopic series of something like that (hint)
$endgroup$
– DonAntonio
Jan 24 at 23:01
$begingroup$
You divided your convergent series into three divergent ones. Doesn't look very promising...unless you can produce a telescopic series of something like that (hint)
$endgroup$
– DonAntonio
Jan 24 at 23:01
$begingroup$
You divided your convergent series into three divergent ones. Doesn't look very promising...unless you can produce a telescopic series of something like that (hint)
$endgroup$
– DonAntonio
Jan 24 at 23:01
add a comment |
3 Answers
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Hint: $frac {n+1} {n(2n-1)(2n+1)}=frac 3 4 frac 1 {n(2n-1)} -frac 1 4 frac 1 {n(2n+1)}$. So what is left is to find the sums $sum frac 1 {n(2n-1)}$ and $sum frac 1 {n(2n+1)}$.
answered Jan 24 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68k53068
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Proceeding on the partial fraction decomposition you already got
$$
eqalign{
& S = sumlimits_{2, le ,n} {{{n + 1} over {nleft( {2n - 1} right)left( {2n + 1} right)}}} = cr
& = sumlimits_{2, le ,n} {{{1/2} over {left( {2n + 1} right)}} + {{3/2} over {left( {2n - 1} right)}} + {1 over n}} = cr
& = sumlimits_{0, le ,n} {{{1/2} over {left( {2n + 5} right)}} + {{3/2} over {left( {2n + 3} right)}} - {1 over {n + 2}}} = cr
& = sumlimits_{0, le ,n} {{{1/4} over {left( {n + 5/2} right)}} + {{3/4} over {left( {n + 3/2} right)}} - {1 over {n + 2}}} cr}
$$
Since
$$
sumnolimits_{n = 0}^N {{1 over {n + a}}} = sumnolimits_{n = a}^{N + a} {{1 over n}} = psi (N + a) - psi (a)
$$
then
$$
eqalign{
& S = mathop {lim }limits_{N, to ,infty } left( matrix{
{1 over 4}psi (N + 5/2) + {3 over 4}psi (N + 3/2) - psi (N + 2) + hfill cr
+ psi (2) - {1 over 4}psi (5/2) - {3 over 4}psi (3/2) hfill cr} right) = cr
& = psi (2) - {1 over 4}psi (5/2) - {3 over 4}psi (3/2) = cr
& = 1 - gamma - {1 over 4}left( {{8 over 3} - 2ln 2 - gamma } right) - {3 over 4}left( {2 - 2ln 2 - gamma } right) = cr
& = 2ln 2 - {7 over 6} cr}
$$
answered Jan 24 at 23:46
G CabG Cab
20.2k31340
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add a comment |
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Here is a way to find those two sums @Kavi Rama Murthy refers to in his answer. They are found by first converting them to double integrals.
Let
$$S = sum_{n = 2}^infty frac{n + 1}{n(2n - 1)(2n + 1)} = frac{3}{4} sum_{n = 2}^infty frac{1}{n (2n - 1)} - frac{1}{4} sum_{n = 1}^infty frac{1}{n(2n + 1)} + frac{1}{12}. qquad (*)$$
For the first of the sums appearing to the right in ($*$), observe that
$$frac{1}{n} = int_0^1 x^{n - 1} , dx quad text{and} quad frac{1}{2n - 1} = int_0^1 y^{2n - 2} , dy.$$
The sum may now be rewritten as
begin{align}
sum_{n = 2}^infty frac{1}{n(2n - 1)} &= sum_{n = 2}^infty int_0^1 int_0^1 x^{n - 1} y^{2n - 2} , dx , dy\
&= int_0^1 int_0^1 frac{1}{xy^2} sum_{n = 2}^infty (xy^2)^n , dx ,dy tag1\
&= int_0^1 int_0^1 frac{1}{xy^2} cdot frac{(xy^2)^2}{1 - xy^2} , dx , dy tag2\
&= int_0^1 frac{1}{y^2} int_0^{y^2} frac{u}{1 - u} , du , dy tag3\
&= -int_0^1 frac{1}{y^2} int_0^{y^2} left (1 - frac{1}{1 - u} right ) , du , dy\
&= -int_0^1 left (1 + frac{ln (1 - y^2)}{y^2} right ) , dy\
&= -1 + 2 int_0^1 frac{1 - y}{1 - y^2} , dy tag4\
&= -1 + 2 int_0^1 frac{dy}{1 + y}\
&= -1 + 2 ln 2.
end{align}
Explanation
(1) Interchange of the sum and integral signs is justified by the dominated convergence theorem.
(2) Summing of a geometric series.
(3) Change of variable $u = xy^2$ is made.
(4) Integration by parts has been used.
The second of the sums appearing in ($*$) can be found in a similar manner (see here). The result is:
$$sum_{n = 1}^infty frac{1}{n(2n + 1)} = 2 - 2ln 2.$$
Thus
$$S = frac{3}{4} (2 ln 2 - 1) - frac{1}{4} (2 - 2 ln 2) + frac{1}{12},$$
or
$$sum_{n = 2}^infty frac{n + 1}{n(2n - 1)(2n + 1)} = 2 ln 2 - frac{7}{6}.$$
answered Jan 25 at 5:41
omegadotomegadot
6,4072829
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another approach to confirm the correctness of my answer: thanks.
$endgroup$
– G Cab
Jan 26 at 19:26
add a comment |
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3 Answers
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3 Answers
3
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$begingroup$
Hint: $frac {n+1} {n(2n-1)(2n+1)}=frac 3 4 frac 1 {n(2n-1)} -frac 1 4 frac 1 {n(2n+1)}$. So what is left is to find the sums $sum frac 1 {n(2n-1)}$ and $sum frac 1 {n(2n+1)}$.
answered Jan 24 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68k53068
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add a comment |
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Hint: $frac {n+1} {n(2n-1)(2n+1)}=frac 3 4 frac 1 {n(2n-1)} -frac 1 4 frac 1 {n(2n+1)}$. So what is left is to find the sums $sum frac 1 {n(2n-1)}$ and $sum frac 1 {n(2n+1)}$.
answered Jan 24 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68k53068
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add a comment |
$begingroup$
Hint: $frac {n+1} {n(2n-1)(2n+1)}=frac 3 4 frac 1 {n(2n-1)} -frac 1 4 frac 1 {n(2n+1)}$. So what is left is to find the sums $sum frac 1 {n(2n-1)}$ and $sum frac 1 {n(2n+1)}$.
answered Jan 24 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68k53068
$endgroup$
Hint: $frac {n+1} {n(2n-1)(2n+1)}=frac 3 4 frac 1 {n(2n-1)} -frac 1 4 frac 1 {n(2n+1)}$. So what is left is to find the sums $sum frac 1 {n(2n-1)}$ and $sum frac 1 {n(2n+1)}$.
answered Jan 24 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68k53068
answered Jan 24 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68k53068
answered Jan 24 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68k53068
answered Jan 24 at 23:28
Kavi Rama MurthyKavi Rama Murthy
68k53068
68k53068
add a comment |
add a comment |
$begingroup$
Proceeding on the partial fraction decomposition you already got
$$
eqalign{
& S = sumlimits_{2, le ,n} {{{n + 1} over {nleft( {2n - 1} right)left( {2n + 1} right)}}} = cr
& = sumlimits_{2, le ,n} {{{1/2} over {left( {2n + 1} right)}} + {{3/2} over {left( {2n - 1} right)}} + {1 over n}} = cr
& = sumlimits_{0, le ,n} {{{1/2} over {left( {2n + 5} right)}} + {{3/2} over {left( {2n + 3} right)}} - {1 over {n + 2}}} = cr
& = sumlimits_{0, le ,n} {{{1/4} over {left( {n + 5/2} right)}} + {{3/4} over {left( {n + 3/2} right)}} - {1 over {n + 2}}} cr}
$$
Since
$$
sumnolimits_{n = 0}^N {{1 over {n + a}}} = sumnolimits_{n = a}^{N + a} {{1 over n}} = psi (N + a) - psi (a)
$$
then
$$
eqalign{
& S = mathop {lim }limits_{N, to ,infty } left( matrix{
{1 over 4}psi (N + 5/2) + {3 over 4}psi (N + 3/2) - psi (N + 2) + hfill cr
+ psi (2) - {1 over 4}psi (5/2) - {3 over 4}psi (3/2) hfill cr} right) = cr
& = psi (2) - {1 over 4}psi (5/2) - {3 over 4}psi (3/2) = cr
& = 1 - gamma - {1 over 4}left( {{8 over 3} - 2ln 2 - gamma } right) - {3 over 4}left( {2 - 2ln 2 - gamma } right) = cr
& = 2ln 2 - {7 over 6} cr}
$$
answered Jan 24 at 23:46
G CabG Cab
20.2k31340
$endgroup$
add a comment |
$begingroup$
Proceeding on the partial fraction decomposition you already got
$$
eqalign{
& S = sumlimits_{2, le ,n} {{{n + 1} over {nleft( {2n - 1} right)left( {2n + 1} right)}}} = cr
& = sumlimits_{2, le ,n} {{{1/2} over {left( {2n + 1} right)}} + {{3/2} over {left( {2n - 1} right)}} + {1 over n}} = cr
& = sumlimits_{0, le ,n} {{{1/2} over {left( {2n + 5} right)}} + {{3/2} over {left( {2n + 3} right)}} - {1 over {n + 2}}} = cr
& = sumlimits_{0, le ,n} {{{1/4} over {left( {n + 5/2} right)}} + {{3/4} over {left( {n + 3/2} right)}} - {1 over {n + 2}}} cr}
$$
Since
$$
sumnolimits_{n = 0}^N {{1 over {n + a}}} = sumnolimits_{n = a}^{N + a} {{1 over n}} = psi (N + a) - psi (a)
$$
then
$$
eqalign{
& S = mathop {lim }limits_{N, to ,infty } left( matrix{
{1 over 4}psi (N + 5/2) + {3 over 4}psi (N + 3/2) - psi (N + 2) + hfill cr
+ psi (2) - {1 over 4}psi (5/2) - {3 over 4}psi (3/2) hfill cr} right) = cr
& = psi (2) - {1 over 4}psi (5/2) - {3 over 4}psi (3/2) = cr
& = 1 - gamma - {1 over 4}left( {{8 over 3} - 2ln 2 - gamma } right) - {3 over 4}left( {2 - 2ln 2 - gamma } right) = cr
& = 2ln 2 - {7 over 6} cr}
$$
answered Jan 24 at 23:46
G CabG Cab
20.2k31340
$endgroup$
add a comment |
$begingroup$
Proceeding on the partial fraction decomposition you already got
$$
eqalign{
& S = sumlimits_{2, le ,n} {{{n + 1} over {nleft( {2n - 1} right)left( {2n + 1} right)}}} = cr
& = sumlimits_{2, le ,n} {{{1/2} over {left( {2n + 1} right)}} + {{3/2} over {left( {2n - 1} right)}} + {1 over n}} = cr
& = sumlimits_{0, le ,n} {{{1/2} over {left( {2n + 5} right)}} + {{3/2} over {left( {2n + 3} right)}} - {1 over {n + 2}}} = cr
& = sumlimits_{0, le ,n} {{{1/4} over {left( {n + 5/2} right)}} + {{3/4} over {left( {n + 3/2} right)}} - {1 over {n + 2}}} cr}
$$
Since
$$
sumnolimits_{n = 0}^N {{1 over {n + a}}} = sumnolimits_{n = a}^{N + a} {{1 over n}} = psi (N + a) - psi (a)
$$
then
$$
eqalign{
& S = mathop {lim }limits_{N, to ,infty } left( matrix{
{1 over 4}psi (N + 5/2) + {3 over 4}psi (N + 3/2) - psi (N + 2) + hfill cr
+ psi (2) - {1 over 4}psi (5/2) - {3 over 4}psi (3/2) hfill cr} right) = cr
& = psi (2) - {1 over 4}psi (5/2) - {3 over 4}psi (3/2) = cr
& = 1 - gamma - {1 over 4}left( {{8 over 3} - 2ln 2 - gamma } right) - {3 over 4}left( {2 - 2ln 2 - gamma } right) = cr
& = 2ln 2 - {7 over 6} cr}
$$
answered Jan 24 at 23:46
G CabG Cab
20.2k31340
$endgroup$
Proceeding on the partial fraction decomposition you already got
$$
eqalign{
& S = sumlimits_{2, le ,n} {{{n + 1} over {nleft( {2n - 1} right)left( {2n + 1} right)}}} = cr
& = sumlimits_{2, le ,n} {{{1/2} over {left( {2n + 1} right)}} + {{3/2} over {left( {2n - 1} right)}} + {1 over n}} = cr
& = sumlimits_{0, le ,n} {{{1/2} over {left( {2n + 5} right)}} + {{3/2} over {left( {2n + 3} right)}} - {1 over {n + 2}}} = cr
& = sumlimits_{0, le ,n} {{{1/4} over {left( {n + 5/2} right)}} + {{3/4} over {left( {n + 3/2} right)}} - {1 over {n + 2}}} cr}
$$
Since
$$
sumnolimits_{n = 0}^N {{1 over {n + a}}} = sumnolimits_{n = a}^{N + a} {{1 over n}} = psi (N + a) - psi (a)
$$
then
$$
eqalign{
& S = mathop {lim }limits_{N, to ,infty } left( matrix{
{1 over 4}psi (N + 5/2) + {3 over 4}psi (N + 3/2) - psi (N + 2) + hfill cr
+ psi (2) - {1 over 4}psi (5/2) - {3 over 4}psi (3/2) hfill cr} right) = cr
& = psi (2) - {1 over 4}psi (5/2) - {3 over 4}psi (3/2) = cr
& = 1 - gamma - {1 over 4}left( {{8 over 3} - 2ln 2 - gamma } right) - {3 over 4}left( {2 - 2ln 2 - gamma } right) = cr
& = 2ln 2 - {7 over 6} cr}
$$
answered Jan 24 at 23:46
G CabG Cab
20.2k31340
answered Jan 24 at 23:46
G CabG Cab
20.2k31340
answered Jan 24 at 23:46
G CabG Cab
20.2k31340
answered Jan 24 at 23:46
G CabG Cab
20.2k31340
20.2k31340
add a comment |
add a comment |
$begingroup$
Here is a way to find those two sums @Kavi Rama Murthy refers to in his answer. They are found by first converting them to double integrals.
Let
$$S = sum_{n = 2}^infty frac{n + 1}{n(2n - 1)(2n + 1)} = frac{3}{4} sum_{n = 2}^infty frac{1}{n (2n - 1)} - frac{1}{4} sum_{n = 1}^infty frac{1}{n(2n + 1)} + frac{1}{12}. qquad (*)$$
For the first of the sums appearing to the right in ($*$), observe that
$$frac{1}{n} = int_0^1 x^{n - 1} , dx quad text{and} quad frac{1}{2n - 1} = int_0^1 y^{2n - 2} , dy.$$
The sum may now be rewritten as
begin{align}
sum_{n = 2}^infty frac{1}{n(2n - 1)} &= sum_{n = 2}^infty int_0^1 int_0^1 x^{n - 1} y^{2n - 2} , dx , dy\
&= int_0^1 int_0^1 frac{1}{xy^2} sum_{n = 2}^infty (xy^2)^n , dx ,dy tag1\
&= int_0^1 int_0^1 frac{1}{xy^2} cdot frac{(xy^2)^2}{1 - xy^2} , dx , dy tag2\
&= int_0^1 frac{1}{y^2} int_0^{y^2} frac{u}{1 - u} , du , dy tag3\
&= -int_0^1 frac{1}{y^2} int_0^{y^2} left (1 - frac{1}{1 - u} right ) , du , dy\
&= -int_0^1 left (1 + frac{ln (1 - y^2)}{y^2} right ) , dy\
&= -1 + 2 int_0^1 frac{1 - y}{1 - y^2} , dy tag4\
&= -1 + 2 int_0^1 frac{dy}{1 + y}\
&= -1 + 2 ln 2.
end{align}
Explanation
(1) Interchange of the sum and integral signs is justified by the dominated convergence theorem.
(2) Summing of a geometric series.
(3) Change of variable $u = xy^2$ is made.
(4) Integration by parts has been used.
The second of the sums appearing in ($*$) can be found in a similar manner (see here). The result is:
$$sum_{n = 1}^infty frac{1}{n(2n + 1)} = 2 - 2ln 2.$$
Thus
$$S = frac{3}{4} (2 ln 2 - 1) - frac{1}{4} (2 - 2 ln 2) + frac{1}{12},$$
or
$$sum_{n = 2}^infty frac{n + 1}{n(2n - 1)(2n + 1)} = 2 ln 2 - frac{7}{6}.$$
answered Jan 25 at 5:41
omegadotomegadot
6,4072829
$endgroup$
$begingroup$
another approach to confirm the correctness of my answer: thanks.
$endgroup$
– G Cab
Jan 26 at 19:26
add a comment |
$begingroup$
Here is a way to find those two sums @Kavi Rama Murthy refers to in his answer. They are found by first converting them to double integrals.
Let
$$S = sum_{n = 2}^infty frac{n + 1}{n(2n - 1)(2n + 1)} = frac{3}{4} sum_{n = 2}^infty frac{1}{n (2n - 1)} - frac{1}{4} sum_{n = 1}^infty frac{1}{n(2n + 1)} + frac{1}{12}. qquad (*)$$
For the first of the sums appearing to the right in ($*$), observe that
$$frac{1}{n} = int_0^1 x^{n - 1} , dx quad text{and} quad frac{1}{2n - 1} = int_0^1 y^{2n - 2} , dy.$$
The sum may now be rewritten as
begin{align}
sum_{n = 2}^infty frac{1}{n(2n - 1)} &= sum_{n = 2}^infty int_0^1 int_0^1 x^{n - 1} y^{2n - 2} , dx , dy\
&= int_0^1 int_0^1 frac{1}{xy^2} sum_{n = 2}^infty (xy^2)^n , dx ,dy tag1\
&= int_0^1 int_0^1 frac{1}{xy^2} cdot frac{(xy^2)^2}{1 - xy^2} , dx , dy tag2\
&= int_0^1 frac{1}{y^2} int_0^{y^2} frac{u}{1 - u} , du , dy tag3\
&= -int_0^1 frac{1}{y^2} int_0^{y^2} left (1 - frac{1}{1 - u} right ) , du , dy\
&= -int_0^1 left (1 + frac{ln (1 - y^2)}{y^2} right ) , dy\
&= -1 + 2 int_0^1 frac{1 - y}{1 - y^2} , dy tag4\
&= -1 + 2 int_0^1 frac{dy}{1 + y}\
&= -1 + 2 ln 2.
end{align}
Explanation
(1) Interchange of the sum and integral signs is justified by the dominated convergence theorem.
(2) Summing of a geometric series.
(3) Change of variable $u = xy^2$ is made.
(4) Integration by parts has been used.
The second of the sums appearing in ($*$) can be found in a similar manner (see here). The result is:
$$sum_{n = 1}^infty frac{1}{n(2n + 1)} = 2 - 2ln 2.$$
Thus
$$S = frac{3}{4} (2 ln 2 - 1) - frac{1}{4} (2 - 2 ln 2) + frac{1}{12},$$
or
$$sum_{n = 2}^infty frac{n + 1}{n(2n - 1)(2n + 1)} = 2 ln 2 - frac{7}{6}.$$
answered Jan 25 at 5:41
omegadotomegadot
6,4072829
$endgroup$
$begingroup$
another approach to confirm the correctness of my answer: thanks.
$endgroup$
– G Cab
Jan 26 at 19:26
add a comment |
$begingroup$
Here is a way to find those two sums @Kavi Rama Murthy refers to in his answer. They are found by first converting them to double integrals.
Let
$$S = sum_{n = 2}^infty frac{n + 1}{n(2n - 1)(2n + 1)} = frac{3}{4} sum_{n = 2}^infty frac{1}{n (2n - 1)} - frac{1}{4} sum_{n = 1}^infty frac{1}{n(2n + 1)} + frac{1}{12}. qquad (*)$$
For the first of the sums appearing to the right in ($*$), observe that
$$frac{1}{n} = int_0^1 x^{n - 1} , dx quad text{and} quad frac{1}{2n - 1} = int_0^1 y^{2n - 2} , dy.$$
The sum may now be rewritten as
begin{align}
sum_{n = 2}^infty frac{1}{n(2n - 1)} &= sum_{n = 2}^infty int_0^1 int_0^1 x^{n - 1} y^{2n - 2} , dx , dy\
&= int_0^1 int_0^1 frac{1}{xy^2} sum_{n = 2}^infty (xy^2)^n , dx ,dy tag1\
&= int_0^1 int_0^1 frac{1}{xy^2} cdot frac{(xy^2)^2}{1 - xy^2} , dx , dy tag2\
&= int_0^1 frac{1}{y^2} int_0^{y^2} frac{u}{1 - u} , du , dy tag3\
&= -int_0^1 frac{1}{y^2} int_0^{y^2} left (1 - frac{1}{1 - u} right ) , du , dy\
&= -int_0^1 left (1 + frac{ln (1 - y^2)}{y^2} right ) , dy\
&= -1 + 2 int_0^1 frac{1 - y}{1 - y^2} , dy tag4\
&= -1 + 2 int_0^1 frac{dy}{1 + y}\
&= -1 + 2 ln 2.
end{align}
Explanation
(1) Interchange of the sum and integral signs is justified by the dominated convergence theorem.
(2) Summing of a geometric series.
(3) Change of variable $u = xy^2$ is made.
(4) Integration by parts has been used.
The second of the sums appearing in ($*$) can be found in a similar manner (see here). The result is:
$$sum_{n = 1}^infty frac{1}{n(2n + 1)} = 2 - 2ln 2.$$
Thus
$$S = frac{3}{4} (2 ln 2 - 1) - frac{1}{4} (2 - 2 ln 2) + frac{1}{12},$$
or
$$sum_{n = 2}^infty frac{n + 1}{n(2n - 1)(2n + 1)} = 2 ln 2 - frac{7}{6}.$$
answered Jan 25 at 5:41
omegadotomegadot
6,4072829
$endgroup$
Here is a way to find those two sums @Kavi Rama Murthy refers to in his answer. They are found by first converting them to double integrals.
Let
$$S = sum_{n = 2}^infty frac{n + 1}{n(2n - 1)(2n + 1)} = frac{3}{4} sum_{n = 2}^infty frac{1}{n (2n - 1)} - frac{1}{4} sum_{n = 1}^infty frac{1}{n(2n + 1)} + frac{1}{12}. qquad (*)$$
For the first of the sums appearing to the right in ($*$), observe that
$$frac{1}{n} = int_0^1 x^{n - 1} , dx quad text{and} quad frac{1}{2n - 1} = int_0^1 y^{2n - 2} , dy.$$
The sum may now be rewritten as
begin{align}
sum_{n = 2}^infty frac{1}{n(2n - 1)} &= sum_{n = 2}^infty int_0^1 int_0^1 x^{n - 1} y^{2n - 2} , dx , dy\
&= int_0^1 int_0^1 frac{1}{xy^2} sum_{n = 2}^infty (xy^2)^n , dx ,dy tag1\
&= int_0^1 int_0^1 frac{1}{xy^2} cdot frac{(xy^2)^2}{1 - xy^2} , dx , dy tag2\
&= int_0^1 frac{1}{y^2} int_0^{y^2} frac{u}{1 - u} , du , dy tag3\
&= -int_0^1 frac{1}{y^2} int_0^{y^2} left (1 - frac{1}{1 - u} right ) , du , dy\
&= -int_0^1 left (1 + frac{ln (1 - y^2)}{y^2} right ) , dy\
&= -1 + 2 int_0^1 frac{1 - y}{1 - y^2} , dy tag4\
&= -1 + 2 int_0^1 frac{dy}{1 + y}\
&= -1 + 2 ln 2.
end{align}
Explanation
(1) Interchange of the sum and integral signs is justified by the dominated convergence theorem.
(2) Summing of a geometric series.
(3) Change of variable $u = xy^2$ is made.
(4) Integration by parts has been used.
The second of the sums appearing in ($*$) can be found in a similar manner (see here). The result is:
$$sum_{n = 1}^infty frac{1}{n(2n + 1)} = 2 - 2ln 2.$$
Thus
$$S = frac{3}{4} (2 ln 2 - 1) - frac{1}{4} (2 - 2 ln 2) + frac{1}{12},$$
or
$$sum_{n = 2}^infty frac{n + 1}{n(2n - 1)(2n + 1)} = 2 ln 2 - frac{7}{6}.$$
answered Jan 25 at 5:41
omegadotomegadot
6,4072829
answered Jan 25 at 5:41
omegadotomegadot
6,4072829
answered Jan 25 at 5:41
omegadotomegadot
6,4072829
answered Jan 25 at 5:41
omegadotomegadot
6,4072829
6,4072829
$begingroup$
another approach to confirm the correctness of my answer: thanks.
$endgroup$
– G Cab
Jan 26 at 19:26
add a comment |
$begingroup$
another approach to confirm the correctness of my answer: thanks.
$endgroup$
– G Cab
Jan 26 at 19:26
$begingroup$
another approach to confirm the correctness of my answer: thanks.
$endgroup$
– G Cab
Jan 26 at 19:26
$begingroup$
another approach to confirm the correctness of my answer: thanks.
$endgroup$
– G Cab
Jan 26 at 19:26
add a comment |
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You divided your convergent series into three divergent ones. Doesn't look very promising...unless you can produce a telescopic series of something like that (hint)
$endgroup$
– DonAntonio
Jan 24 at 23:01