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Let \(\phi \) denote the golden ratio, \(\phi = \dfrac{1+\sqrt5}2 \), prove that \[\phi -\frac { 13 }{ 8 } =\displaystyle\sum _{ n=0 }^{ \infty }{ \frac { (-1)^{ n+1 }(2n+1)! }{ (n+2)!n!{ 4 }^{ 2n+3 } } } . \]

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Lemma :

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

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

Proof :

Using Infinite Binomial Series,

\(\displaystyle (1+x)^{-n} = \sum_{r=0}^{\infty} (-1)^{r} \dfrac{\Gamma(n+r)}{\Gamma(n)\Gamma(r+1)} x^r\)

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

\(\displaystyle f(1) - f(2) = \Gamma\left(\dfrac{3}{2}\right) \int_{0}^{1} (1-x) \left(1+\dfrac{x}{4}\right)^{-\frac{3}{2}} \mathrm{d}x\)

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

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

This proves the Lemma.

Now, the series can be written as,

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

Using Gamma Duplication Formula, we have,

\(\displaystyle \text{S} = -\dfrac{1}{2^5 \sqrt{\pi}} \sum_{r=0}^{\infty} (-1)^{r} \dfrac{\Gamma(r+\frac{3}{2})}{4^r \Gamma(r+3)} \)

\(\displaystyle = -\dfrac{1}{2^5 \sqrt{\pi}} \sum_{r=0}^{\infty} (-1)^{r} \dfrac{ \Gamma(r+\frac{3}{2})}{4^r (r+1)(r+2) \Gamma(r+1)}\)

\(\displaystyle = -\dfrac{1}{2^5 \sqrt{\pi}} \left[\sum_{r=0}^{\infty} (-1)^{r} \dfrac{ \Gamma(r+\frac{3}{2})}{4^r (r+1) \Gamma(r+1)} -\sum_{r=0}^{\infty} (-1)^{r} \dfrac{\Gamma(r+\frac{3}{2})}{4^r (r+2) \Gamma(r+1)} \right]\)

Using the Lemma,

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

Simplifying, we have,

\(\displaystyle \boxed{\text{S} = \phi -\dfrac{13}{8}}\)

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

@Ishan Singh

you're the best at these problems :)

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## Comments

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TopNewestLemma :If \(\displaystyle f(a) = \sum_{r=0}^{\infty} (-1)^{r} \dfrac{\Gamma\left(r+\frac{3}{2}\right)}{4^r(r+a)\Gamma(r+1)}\)

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

Proof :Using Infinite Binomial Series,

\(\displaystyle (1+x)^{-n} = \sum_{r=0}^{\infty} (-1)^{r} \dfrac{\Gamma(n+r)}{\Gamma(n)\Gamma(r+1)} x^r\)

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

\(\displaystyle f(1) - f(2) = \Gamma\left(\dfrac{3}{2}\right) \int_{0}^{1} (1-x) \left(1+\dfrac{x}{4}\right)^{-\frac{3}{2}} \mathrm{d}x\)

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

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

This proves the Lemma.

Now, the series can be written as,

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

Using Gamma Duplication Formula, we have,

\(\displaystyle \text{S} = -\dfrac{1}{2^5 \sqrt{\pi}} \sum_{r=0}^{\infty} (-1)^{r} \dfrac{\Gamma(r+\frac{3}{2})}{4^r \Gamma(r+3)} \)

\(\displaystyle = -\dfrac{1}{2^5 \sqrt{\pi}} \sum_{r=0}^{\infty} (-1)^{r} \dfrac{ \Gamma(r+\frac{3}{2})}{4^r (r+1)(r+2) \Gamma(r+1)}\)

\(\displaystyle = -\dfrac{1}{2^5 \sqrt{\pi}} \left[\sum_{r=0}^{\infty} (-1)^{r} \dfrac{ \Gamma(r+\frac{3}{2})}{4^r (r+1) \Gamma(r+1)} -\sum_{r=0}^{\infty} (-1)^{r} \dfrac{\Gamma(r+\frac{3}{2})}{4^r (r+2) \Gamma(r+1)} \right]\)

Using the Lemma,

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

Simplifying, we have,

\(\displaystyle \boxed{\text{S} = \phi -\dfrac{13}{8}}\)

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

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@Ishan Singh

you're the best at these problems :)

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