Verified answer

Topic:

Sequence, series, and infinite product.

Some derivations:

We need to prove that,

[tex]{\displaystyle \sum_{n=1}^\infty\dfrac{1}{S_n} = \dfrac{\pi}{2\sqrt{3}}\tan\left(\dfrac{\pi}{2\sqrt3}\right)\text{Where $S_n$ is star number.}}[/tex]

Since the RHS is in the form of trigonometric function tangent, we need to use partial fraction expansion of tan(x).

We know the partial fraction expansion of cos(x):

[tex]\displaystyle \cos(x) = \prod_{n=1}^\infty\left[1-\dfrac{4x^2}{\pi^2 (2n-1)^2}\right][/tex]

Taking log on both sides,

[tex]\displaystyle \log\cos(x) = \sum_{n=1}^\infty\log\left[1-\dfrac{4x^2}{\pi^2 (2n-1)^2}\right][/tex]

[tex]{\displaystyle \log\cos(x) = \sum_{n=1}^\infty\log\left[\dfrac{\pi^2 (2n-1)^2-4x^2}{\pi^2 (2n-1)^2}\right]}[/tex]

Differentiating both sides,

[tex]{\displaystyle \dfrac{d}{dx}\bigg(\log\cos(x)\bigg) = \sum_{n=1}^\infty\dfrac{d}{dx}\left\{\log\left[\dfrac{\pi^2 (2n-1)^2-4x^2}{\pi^2 (2n-1)^2}\right]\right\}}[/tex]

[tex]{\displaystyle -\tan(x) = \sum_{n=1}^\infty\left[\dfrac{\pi^2(2n-1)^2}{\pi^2(2n-1)^2 - 4x^2}\times \dfrac{-8x}{\pi^2(2n-1)^2} \right]}[/tex]

[tex]{\displaystyle -\tan(x) = \sum_{n=1}^\infty\left[\dfrac{-8x}{\pi^2(2n-1)^2 - 4x^2}\right]}[/tex]

[tex]{\displaystyle \tan(x) = \sum_{n=1}^\infty\left[\dfrac{8x}{\pi^2(2n-1)^2 - 4x^2}\right]}[/tex]

Put x = π x

[tex]{\displaystyle \tan(\pi x) = \sum_{n=1}^\infty\left[\dfrac{8\pi x}{\pi^2(2n-1)^2 - 4\pi x^2}\right]}[/tex]

Multiply both sides with π

[tex]{\displaystyle \pi \tan(\pi x) = \sum_{n=1}^\infty\left[\dfrac{8\pi^2 x}{\pi^2(2n-1)^2 - 4\pi ^2x^2}\right]}[/tex]

[tex]{\displaystyle \pi \tan(\pi x) = \sum_{n=1}^\infty\left[\dfrac{8 x}{(2n-1)^2 - 4 x^2}\right]}[/tex]

[tex]{\displaystyle \pi \tan(\pi x) = \sum_{n=1}^\infty\left[\dfrac{8 x}{4n^2 - 4n + 1 - 4 x^2}\right]}[/tex]

[tex]{\displaystyle \pi \tan(\pi x) = \sum_{n=1}^\infty\left[\dfrac{2x}{n^2 - 1 + \frac14 - x^2}\right]}[/tex]

[tex]{\displaystyle \pi \tan(\pi x) = \sum_{n=1}^\infty\left[\dfrac{2x}{\left(n - \frac{1}{2}\right)^2 - x^2}\right]}[/tex]

[tex]{\displaystyle \pi \tan(\pi x) = 2x\sum_{n=1}^\infty\left[\dfrac{1}{\left(n - \frac{1}{2}\right)^2 - x^2}\right] \equiv f(x)\qquad \text{(Assume)}}[/tex]

We will use this sum to evaluate the given problem.

Solution:

We have,

[tex]{\displaystyle \sum_{n=1}^\infty\dfrac{1}{S_n}\qquad\qquad\text{Where $S_n$ is star number.}}[/tex]

[tex]{\tt\bullet\quad Star\ number\ is\ given\ by\ S_n = 6n(n - 1) + 1}[/tex]

[tex]{\displaystyle\implies \sum_{n=1}^\infty\dfrac{1}{6n(n-1) + 1}}[/tex]

[tex]{\displaystyle \implies\sum_{n=1}^\infty\dfrac{1}{6n^2 - 6n +1}}[/tex]

[tex]{\displaystyle \implies\dfrac{1}{6}\sum_{n=1}^\infty\dfrac{1}{n^2 -n + \frac16}}[/tex]

[tex]{\displaystyle \implies\dfrac{1}{6}\sum_{n=1}^\infty\dfrac{1}{n^2 -n+ \frac14 - \frac14 + \frac16}}[/tex]

[tex]{\displaystyle \implies\dfrac{1}{6}\sum_{n=1}^\infty\dfrac{1}{\left(n - \frac12\right)^2- \frac1{12}}}[/tex]

[tex]{\displaystyle\implies \dfrac{1}{6}\sum_{n=1}^\infty\dfrac{1}{\left(n - \frac12\right)^2- \left(\frac1{\sqrt{12}} \right)^2}}[/tex]

[tex]{\displaystyle \implies\dfrac{1}{\sqrt{12}} \cdot \left(2 \times \dfrac{1}{\sqrt{12}}\right)\sum_{n=1}^\infty\dfrac{1}{\left(n - \frac12\right)^2- \left(\frac1{\sqrt{12}} \right)^2}}[/tex]

Comparing it with the above-derived function, we get:

[tex]{\displaystyle\implies \dfrac{1}{\sqrt{12}} \cdot f\left(\dfrac1{\sqrt{12}}\right)}[/tex]

[tex]{\displaystyle \implies\dfrac{1}{\sqrt{12}} \cdot\left[\pi \tan\left(\dfrac{\pi}{\sqrt{12}}\right)\right]}[/tex]

[tex]{\displaystyle\implies \dfrac{\pi}{2\sqrt{3}} \tan\left(\dfrac{\pi}{2\sqrt{3}}\right)}[/tex]

Hence we have proved that,

[tex]\boxed{ \sum_{n=1}^\infty\dfrac{1}{S_n} = \dfrac{\pi}{2\sqrt{3}}\tan\left(\dfrac{\pi}{2\sqrt{3}}\right)\qquad\qquad \text{Where $S_n$ is star number.}}[/tex]

Answer:

Hey TakenName, Your Problem had been solved.

Step-by-step explanation:

Since, [tex] \rm{S_{n}} [/tex] is a star number and we know that a star number is a centered hexagram and the [tex] \rm{n^{th}} [/tex] star number is given by: [tex] \rm{S_{n} = 6n(n - 1) + 1} [/tex].

Now, we will take the left hand side of the equation and show that it is equal to the right hand side.

[tex] \rule{180pt}{2pt} [/tex]

Explanation

The digit root of a star number is always 1 or 4 and progresses in the sequence 1,4,1. The last two digits of a star number in base 10 are always

01,13,21,33,37,41,53,61,73,81 or 93.

Unique among the star number is 35113.

We know that,

[tex]\rm{\sum\limits_{n = 1}^{ \infty} \bigg[ \dfrac{1}{S_{n}}\bigg] = \sum\limits_{n = 1}^{ \infty} \bigg[ \dfrac{1}{6n(n - 1) + 1}\bigg]} \\ [/tex]

[tex]\rm{= 1 + \dfrac{1}{13} + \dfrac{1}{37} + \dfrac{1}{73}+\dfrac{1}{121} + \dfrac{1}{181} + ... } \\ [/tex]

[tex]\rm{= \bigg(\dfrac{\Pi}{2\sqrt{3}}\bigg)tan \bigg(\dfrac{\Pi}{2\sqrt{3}}\bigg)}[/tex]

[tex]\rm{≈ 1.159173} \\ [/tex]

Hence, shown!

and it can be seen that the LHS of the question is equal to the RHS.

[tex] \rule{200pt}{3pt} [/tex]