Prove this

Let ϕ\phi denote the golden ratio, ϕ=1+52\phi = \dfrac{1+\sqrt5}2 , prove that ϕ−138=∑n=0∞(−1)n+1(2n+1)!(n+2)!n!42n+3.\phi -\frac { 13 }{ 8 } =\displaystyle\sum _{ n=0 }^{ \infty }{ \frac { (-1)^{ n+1 }(2n+1)! }{ (n+2)!n!{ 4 }^{ 2n+3 } } } .

#Calculus

Note by Hamza A
5 years, 4 months ago

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Comments

Lemma :

If f(a)=∑r=0∞(−1)rΓ(r+32)4r(r+a)Γ(r+1)\displaystyle f(a) = \sum_{r=0}^{\infty} (-1)^{r} \dfrac{\Gamma\left(r+\frac{3}{2}\right)}{4^r(r+a)\Gamma(r+1)}

Then f(1)−f(2)=π2(72−1605)\displaystyle f(1) - f(2) = \dfrac{\sqrt{\pi}}{2} \left(72 - \dfrac{160}{\sqrt{5}}\right)

Proof :

Using Infinite Binomial Series,

(1+x)−n=∑r=0∞(−1)rΓ(n+r)Γ(n)Γ(r+1)xr\displaystyle (1+x)^{-n} = \sum_{r=0}^{\infty} (-1)^{r} \dfrac{\Gamma(n+r)}{\Gamma(n)\Gamma(r+1)} x^r

Putting n=32\displaystyle n = \dfrac{3}{2} and noting that 1z+1=∫01xzdx \displaystyle \dfrac{1}{z+1} = \int_{0}^{1} x^z \mathrm{d}x, we have,

f(1)−f(2)=Γ(32)∫01(1−x)(1+x4)−32dx\displaystyle f(1) - f(2) = \Gamma\left(\dfrac{3}{2}\right) \int_{0}^{1} (1-x) \left(1+\dfrac{x}{4}\right)^{-\frac{3}{2}} \mathrm{d}x

=π2[−16(x+8)x+4]01(∵Γ(32)=π2)\displaystyle = \dfrac{\sqrt{\pi}}{2}\left[\dfrac{-16(x+8)}{\sqrt{x+4}}\right]_{0}^{1} \qquad \left( \because \Gamma \left(\frac{3}{2}\right) = \dfrac{\sqrt{\pi}}{2} \right)

=π2(72−1605)\displaystyle = \dfrac{\sqrt{\pi}}{2} \left(72 - \dfrac{160}{\sqrt{5}}\right)

This proves the Lemma.

Now, the series can be written as,

S=∑r=0∞(−1)r+1Γ(2r+2)42r+3Γ(r+3)Γ(r+1)\displaystyle \text{S} = \sum_{r=0}^{\infty} (-1)^{r+1} \dfrac{ \Gamma(2r+2)}{4^{2r+3} \Gamma(r+3)\Gamma(r+1)}

Using Gamma Duplication Formula, we have,

S=−125π∑r=0∞(−1)rΓ(r+32)4rΓ(r+3)\displaystyle \text{S} = -\dfrac{1}{2^5 \sqrt{\pi}} \sum_{r=0}^{\infty} (-1)^{r} \dfrac{\Gamma(r+\frac{3}{2})}{4^r \Gamma(r+3)}

=−125π∑r=0∞(−1)rΓ(r+32)4r(r+1)(r+2)Γ(r+1)\displaystyle = -\dfrac{1}{2^5 \sqrt{\pi}} \sum_{r=0}^{\infty} (-1)^{r} \dfrac{ \Gamma(r+\frac{3}{2})}{4^r (r+1)(r+2) \Gamma(r+1)}

=−125π[∑r=0∞(−1)rΓ(r+32)4r(r+1)Γ(r+1)−∑r=0∞(−1)rΓ(r+32)4r(r+2)Γ(r+1)]\displaystyle = -\dfrac{1}{2^5 \sqrt{\pi}} \left[\sum_{r=0}^{\infty} (-1)^{r} \dfrac{ \Gamma(r+\frac{3}{2})}{4^r (r+1) \Gamma(r+1)} -\sum_{r=0}^{\infty} (-1)^{r} \dfrac{\Gamma(r+\frac{3}{2})}{4^r (r+2) \Gamma(r+1)} \right]

Using the Lemma,

S=−125π(f(1)−f(2))=−125π(π2(72−1605))\displaystyle \text{S} = -\dfrac{1}{2^5 \sqrt{\pi}} \left(f(1) - f(2)\right) = -\dfrac{1}{2^5 \sqrt{\pi}} \left(\dfrac{\sqrt{\pi}}{2} \left(72 - \dfrac{160}{\sqrt{5}}\right)\right)

Simplifying, we have,

S=ϕ−138\displaystyle \boxed{\text{S} = \phi -\dfrac{13}{8}}

Ishan Singh - 5 years, 3 months ago

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@Hummus a Try this. Little similar to your question.

Ishan Singh - 5 years, 3 months ago

@Ishan Singh

you're the best at these problems :)

Hamza A - 5 years, 3 months ago
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