Double Factorial Sum?

Calculus Level 5

1 2 ⋅ 4 + 1 ⋅ 3 2 ⋅ 4 ⋅ 6 + 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ⋅ 8 + 1 ⋅ 3 ⋅ 5 ⋅ 7 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅ 10 + … = ? \dfrac { 1 }{ 2\cdot 4 } +\dfrac { 1\cdot3 }{ 2\cdot4\cdot6 } +\dfrac { 1\cdot3\cdot5 }{ 2\cdot4\cdot6\cdot8 } +\dfrac { 1\cdot3\cdot5\cdot7 }{ 2\cdot4\cdot6\cdot8\cdot10 } + \ldots = \ ?


The answer is 0.5.

View wiki
Pankaj Joshi by Pankaj Joshi , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

5 solutions

David Vaccaro
Jul 25, 2014

Not as stylish as the other solutions but ( 1 − x ) 1 2 = 1 − x 2 − x 2 2.4 − 1.3. x 3 2.4.6 (1-x)^{\frac{1}{2}}=1-\frac{x}{2}-\frac{x^{2}}{2.4}-\frac{1.3.x^{3}}{2.4.6}

Put x = 1 x=1 gives 0 = 1 − 1 2 − S 0=1-\frac{1}{2}-S and S = 1 2 S=\boxed{\frac{1}{2}} .

55 Helpful 0 Interesting 0 Brilliant 0 Confused

I didn't know of that identity.

Kartik Sharma - 6 years, 4 months ago

You have got the best solution among the lot. Please tell how you thought of it

Ronak Agarwal - 6 years, 10 months ago

Log in to reply

I don't know if this is valid, but are you allowed to partial fraction/split up each fraction.

1 8 = 1 4 − 1 8 \frac{1}{8}=\frac{1}{4}-\frac{1}{8}

1 16 = 1 4 − 3 16 \frac{1}{16}=\frac{1}{4}-\frac{3}{16}

5 128 = 3 16 − 19 128 \frac{5}{128}=\frac{3}{16}-\frac{19}{128}

7 256 = 19 128 − 31 256 \frac{7}{256}=\frac{19}{128}-\frac{31}{256}

21 1024 = 31 256 − 103 1024 \frac{21}{1024}=\frac{31}{256}-\frac{103}{1024}

33 2048 = 103 1024 − 173 2048 \frac{33}{2048}=\frac{103}{1024}-\frac{173}{2048}

429 32768 = 173 2048 − 2339 32768 \frac{429}{32768}=\frac{173}{2048}-\frac{2339}{32768}

Note how everything is of the form a 2 n = b 2 p − a − b ( 2 n − p ) 2 n \frac{a}{2^n}=\frac{b}{2^p}-\frac{a-b(2^{n-p})}{2^n} .

The RHS of the RHS tends 0 farther and farther down the list. Since the n+1 term's LHS of the RHS is the same as n's RHS of the RHS (sorry for my horrid wording), the LHS of the RHS will also tend 0.

So we are left with the first two terms of 1 4 + 1 4 = 1 2 \frac{1}{4}+\frac{1}{4}=\frac{1}{2}

Cool thing to note, the numerator of the second term on theRHSis always a prime.

Trevor Arashiro - 6 years, 4 months ago

Log in to reply

Oh my gosh, this is a really neat sequence of primes.

Now I realize that this is due to the goldbach conjecture.

Trevor Arashiro - 6 years, 4 months ago

I don't get it, what does the the dot (".") in "2.4.6" mean?

DPK ­ - 6 years, 10 months ago

Log in to reply

It means multiply

Trevor Arashiro - 6 years, 4 months ago

Nope, your identity is wrong

( 1 − x ) − n = 1 + ( n 1 ) x + ( n + 1 2 ) x 2 + ( n + 2 3 ) x 3 + . . . (1-x)^{-n} = 1 + {n \choose 1} x + {n + 1 \choose 2}x^2 + {n + 2 \choose 3}x^3 + ...

Its general term is given by (which series you have wrote?, moreover keeping 1/2 does'nt give 2.4 in denominator)

T r + 1 = n ( n + 1 ) ( n + 2 ) . . . . . [ n + ( r − 1 ) ] r ! × x r T_{r+1} = \dfrac{n(n+1)(n+2).....[n+(r-1)]}{r!} \times x^r

S = 1 2.4 + 1.3 2.4.6 + 1.3.5 2.4.6.8 + 1.3.5.7 2.4.6.8.10 + . . . . . ∞ S = \dfrac { 1 }{ 2.4 } +\dfrac { 1.3 }{ 2.4.6 } +\dfrac { 1.3.5 }{ 2.4.6.8 } +\dfrac { 1.3.5.7 }{ 2.4.6.8.10 } +.....\infty

S = 1 1.2 ( 1 2 ) 1 + 1 2 . 3 2 1.2.3 × ( 1 2 ) 2 + 1 2 . 3 2 . 5 2 1.2.3.4 × ( 1 2 ) 3 + . . . . . . . S = \dfrac{1}{1.2}(\dfrac{1}{2})^1 + \dfrac{\dfrac{1}{2}.\dfrac{3}{2}}{1.2.3}\times(\dfrac{1}{2})^2 + \dfrac{\dfrac{1}{2}.\dfrac{3}{2}.\dfrac{5}{2}}{1.2.3.4}\times(\dfrac{1}{2})^3 + .......

S 2 = 1 2 ! ( 1 2 ) 2 + 1 2 . ( 1 + 1 2 ) 3 ! × ( 1 2 ) 3 + 1 2 . ( 1 + 1 2 ) . ( 2 + 1 2 4 ! × ( 1 2 ) 4 + . . . . . . . \dfrac{S}{2} = \dfrac{1}{2!}(\dfrac{1}{2})^2 + \dfrac{ \dfrac{1}{2}.(1 + \dfrac{1}{2})}{3!}\times(\dfrac{1}{2})^3 + \dfrac{\dfrac{1}{2}.(1+\dfrac{1}{2}).(2 + \dfrac{1}{2}}{4!}\times(\dfrac{1}{2})^4 + .......

S 2 = ( 1 − 1 2 ) − 1 / 2 = 2 − 1 − 1 2 = 0.414 − 0.5 × 2 = 0.818 − 1 = − 0.182 \dfrac{S}{2} = ( 1 - \dfrac{1}{2})^{-1/2} = \sqrt{2} - 1 - \dfrac{1}{2} = 0.414 - 0.5\times 2 = 0.818 - 1 = -0.182

While writing answer I made many mistakes

U Z - 6 years, 5 months ago

Log in to reply

Thanks for the solution.

atreyee saha - 5 years, 11 months ago

The identity is right.

Kishore S. Shenoy - 5 years, 9 months ago

From where did you know that identity?????

Sudhir Aripirala - 6 years, 4 months ago

This is amazing

Anirban Mandal - 5 years, 2 months ago
Ronak Agarwal
Jul 24, 2014

W e h a v e t h e s u m a s S = lim ⁡ n → ∞ ∑ r = 1 n 1.3.5........ ( 2 r − 1 ) 2.4.6....... ( 2 r + 2 ) T r = 1.3.5........ ( 2 r − 1 ) 2.4.6....... ( 2 r + 2 ) = 1.3.5........ ( 2 r − 1 ) 2.4.6....... ( 2 r + 2 ) ( ( 2 r + 2 ) − ( 2 r + 1 ) 1 ) ( m u l t i p l y i n g a n d d i v i d i n g b y 1 ) ⇒ T r = 1.3.5........ ( 2 r − 1 ) 2.4.6............ ( 2 r ) − 1.3.5........ ( 2 r + 1 ) 2.4.6....... ( 2 r + 2 ) = V r − V r + 1 ⇒ S = ∑ T r = ∑ V r − V r + 1 = V 1 − lim ⁡ n → ∞ V n = 1 / 2 We\quad have\quad the\quad sum\quad as\quad S=\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { 1.3.5........(2r-1) }{ 2.4.6.......(2r+2) } } } \\ { T }_{ r }=\frac { 1.3.5........(2r-1) }{ 2.4.6.......(2r+2) } =\frac { 1.3.5........(2r-1) }{ 2.4.6.......(2r+2) } (\frac { (2r+2)-(2r+1) }{ 1 } )\quad \\ (multiplying\quad \quad and\quad dividing\quad by\quad 1)\\ \Rightarrow { T }_{ r }=\frac { 1.3.5........(2r-1) }{ 2.4.6............(2r) } -\frac { 1.3.5........(2r+1) }{ 2.4.6.......(2r+2) } ={ V }_{ r }-{ V }_{ r+1 }\\ \Rightarrow S=\sum { { T }_{ r } } =\sum { { V }_{ r }-{ V }_{ r+1 } } ={ V }_{ 1 }-\lim _{ n\rightarrow \infty }{ { V }_{ n } } =1/2

14 Helpful 0 Interesting 0 Brilliant 0 Confused

Excellent! But you need to show that lim ⁡ n → ∞ V n = 0 \lim_{n\to\infty}V_n=0

Vijaysekhar Chellaboina - 6 years, 4 months ago
Michael Mendrin
Jul 15, 2014

By induction, it can be shown that

∑ k = 1 n ( 2 k ) ! 2 k + 1 k ! ( k + 1 ) ! = 1 2 − ( 2 n ) ! 2 n + 1 n ! ( n + 1 ) ! ( 2 n + 1 ) \displaystyle\sum _{ k=1 }^{ n }{ \dfrac { (2k)! }{ { 2 }^{ k+1 }k!(k+1)! } } =\dfrac { 1 }{ 2 } -\dfrac { (2n)! }{ { 2 }^{ n+1 }n!(n+1)! } (2n+1)

so that as n → ∞ n\rightarrow \infty , we end up with 1 2 \dfrac { 1 }{ 2 }
Use Stirling's approximation to find this limit.

See example below to see how this induction works

1 2.4 + 1.3 2.4.6 + 1.3.5 2.4.6.8 = 1 2 − 1.3.5 2.4.6.8 7 \dfrac { 1 }{ 2.4 } +\dfrac { 1.3 }{ 2.4.6 } +\dfrac { 1.3.5 }{ 2.4.6.8 } =\dfrac { 1 }{ 2 } -\dfrac { 1.3.5 }{ 2.4.6.8 } 7

1 2.4 + 1.3 2.4.6 + 1.3.5 2.4.6.8 + 1.3.5.7 2.4.6.8.10 = 1 2 − 1.3.5.7 2.4.6.8.10 9 \dfrac { 1 }{ 2.4 } +\dfrac { 1.3 }{ 2.4.6 } +\dfrac { 1.3.5 }{ 2.4.6.8 } +\dfrac { 1.3.5.7 }{ 2.4.6.8.10 } =\dfrac { 1 }{ 2 } -\dfrac { 1.3.5.7 }{ 2.4.6.8.10 } 9

1.3.5 2.4.6.8 7 − 1.3.5.7 2.4.6.8.10 9 = 1.3.5.7 2.4.6.8.10 \dfrac { 1.3.5 }{ 2.4.6.8 } 7-\dfrac { 1.3.5.7 }{ 2.4.6.8.10 } 9=\dfrac { 1.3.5.7 }{ 2.4.6.8.10 }

6 Helpful 0 Interesting 0 Brilliant 0 Confused

You can use Binomial theorem here, ma'am.

Kishore S. Shenoy - 5 years, 9 months ago

Ma'am? Really?

Sal Gard - 4 years, 10 months ago
Kishore S. Shenoy
Aug 21, 2015

Solution by David Vaccaro clarified

The identity given by David Vaccaro is absolutely right!

S = 1 2 ⋅ 4 x 2 + 1 ⋅ 3 2 ⋅ 4 ⋅ 6 x 3 + 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ⋅ 8 x 4 + ⋯ where x = 1 \displaystyle S = \frac{1}{2\cdot 4}x^2 + \frac{1\cdot 3}{2\cdot 4\cdot 6}x^3 + \frac{1\cdot 3\cdot 5}{2\cdot4\cdot6\cdot8}x^4+ \cdots \text{ where } x = 1\\

According to Binomial Theorem (general),

( 1 − x ) − p q = 1 + p 1 ! x q + p ( p + q ) 2 ! ( x q ) 2 + p ( p + q ) ( p + 2 q ) 3 ! ( x q ) 3 + ⋯ \displaystyle (1-x)^{-\dfrac{p}{q}} = 1 + \frac{p}{1!} \frac{x}{q} + \frac{p(p+q)}{2!}\left(\frac{x}{q}\right)^2 + \frac{p(p+q)(p+2q)}{3!}\left(\frac{x}{q}\right)^3 + \cdots\\

Putting p = − 1 , q = 2 p = -1,~q = 2

( 1 − x ) − p q = 1 + p 1 ! x q + p ( p + q ) 2 ! ( x q ) 2 + p ( p + q ) ( p + 2 q ) 3 ! ( x q ) 3 + ⋯ = ( 1 − x ) 1 2 = 1 − 1 1 ! x 2 − 1 ( − 1 + 2 ) 2 ! ( x 2 ) 2 − 1 ( − 1 + 2 ) ( − 1 + 4 ) 3 ! ( x 2 ) 3 + ⋯ = 1 − 1 1 ! x 2 − 1 ⋅ 1 2 ! ( x 2 ) 2 − 1 ⋅ 1 ⋅ 3 3 ! ( x 2 ) 3 − 1 ⋅ 1 ⋅ 3 ⋅ 5 4 ! ( x 2 ) 4 + ⋯ = 1 − 1 2 x − 1 2 ⋅ 4 x 2 − 1 ⋅ 3 2 ⋅ 4 ⋅ 6 x 3 − 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ⋅ 8 x 4 + ⋯ Hence, when x = 1 ( 1 − 1 ) 1 2 = 0 = 1 − 1 2 − S ∴ S = 1 2 \displaystyle\begin{aligned} (1-x)^{-\dfrac{p}{q}} &= 1 + \frac{p}{1!} \frac{x}{q} + \frac{p(p+q)}{2!}\left(\frac{x}{q}\right)^2 + \frac{p(p+q)(p+2q)}{3!}\left(\frac{x}{q}\right)^3 + \cdots\\ &=(1-x)^{\frac{1}{2}} \\ &= 1 - \frac{1}{1!} \dfrac{x}{2} - \frac{1(-1+2)}{2!}\left(\frac{x}{2}\right)^2 - \frac{1(-1+2)(-1+4)}{3!}\left(\frac{x}{2}\right)^3 + \cdots\\ &= 1 - \frac{1}{1!} \dfrac{x}{2} - \frac{1\cdot 1}{2!}\left(\frac{x}{2}\right)^2 - \frac{1\cdot 1\cdot 3}{3!}\left(\frac{x}{2}\right)^3 - \frac{1\cdot 1\cdot 3\cdot 5}{4!}\left(\frac{x}{2}\right)^4+ \cdots\\ &= 1 - \frac{1}{2}x - \frac{1}{2\cdot 4}x^2 - \frac{1\cdot 3}{2\cdot 4\cdot 6}x^3 - \frac{1\cdot 3\cdot 5}{2\cdot4\cdot6\cdot8}x^4+ \cdots\\\\ \text{ Hence, when } &\normalsize x = 1\\ (1-1)^{\frac{1}{2}} &= 0 \\ &=1 - \dfrac{1}{2} - S \\ \\ \boxed{\therefore S = \dfrac{1}{2}} \end{aligned}

Thanks for this wonderful solution, David Vaccaro!! \large \text{Thanks for this wonderful solution, David Vaccaro!!}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

You have to be careful here. Note that you haven't yet established that the sum converges at x = 1 x = 1 . You will instead need to prove that it converges to a finite value first.

The binomial theorem is valid for ∣ x ∣ < 1 |x| < 1 , and we have to check the case when ∣ x ∣ = 1 |x| = 1 carefully. For example, in the expansion of ( 1 + x ) − 1 ( 1 + x) ^ {-1} , we cannot apply x = 1 x = 1 to conclude that

1 1 + 1 = ( 1 + 1 ) − 1 = 1 − 1 + 1 − 1 + 1 − 1 + … \frac{1}{ 1 + 1 } = ( 1 + 1) ^ { -1 } = 1 -1 + 1 - 1 + 1 - 1 + \ldots

You have to be careful here. Note that you haven't yet established that the sum converges at x = 1 x = 1 . You will instead need to prove that it converges to a finite value first.

The binomial theorem is valid for ∣ x ∣ < 1 |x| < 1 , and we have to check the case when ∣ x ∣ = 1 |x| = 1 carefully. For example, in the expansion of ( 1 + x ) − 1 ( 1 + x) ^ {-1} , we cannot apply x = 1 x = 1 to conclude that

1 1 + 1 = ( 1 + 1 ) − 1 = 1 − 1 + 1 − 1 + 1 − 1 + … \frac{1}{ 1 + 1 } = ( 1 + 1) ^ { -1 } = 1 -1 + 1 - 1 + 1 - 1 + \ldots

Calvin Lin Staff - 5 years, 9 months ago

Log in to reply

I guess, Root test will do that for me @Calvin Lin sir. It gave me some value < 1 < 1 if I am not wrong.

Kishore S. Shenoy - 5 years, 9 months ago

Log in to reply

I'm not certain if the root test works. I doubt it does. Each additional term 2 n − 1 2 n + 2 \frac{2n-1}{2n+2} brings you much closer to the value of 1, and so the limit of the root would be 1. This can be formalized better using Stirling's Formula .

Calvin Lin Staff - 5 years, 9 months ago

Note that the n t h n^\mathrm{th} term in the sum can be recognized as C n 2 n + 1 \frac{C_n}{2^{n+1}} where C n C_n is the nth Catalan number. Thus the sum is 1 / 2 ∑ n ≥ 1 C n ( 1 / 4 ) n = 1 − 1 / 2 = 1 / 2 1/2\sum_{n\ge 1}C_n (1/4)^n=1-1/2=\boxed{1/2}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...