Explained: "What is the value of ∑k=1∞1(3k+1)(3k+2)"

Emerson Barnes

Answered question

2022-01-24

What is the value of

\sum_{k = 1}^{\infty} \frac{1}{(3 k + 1) (3 k + 2)}

Armani Dyer

Beginner2022-01-25Added 10 answers

I like solution with Gamma/Beta function:

\sum_{k = 1}^{+ \infty} \frac{1}{(3 k + 1) (3 k + 2)} = \sum_{k = 1}^{+ \infty} \frac{Γ (3 k + 1)}{Γ (3 k + 3)}

= \sum_{k = 1}^{+ \infty} B (3 k + 1, 2)

= \sum_{k = 1}^{+ \infty} \int_{0}^{1} x^{3 k} (1 - x) d x

= \int_{0}^{1} \sum_{k = 1}^{+ \infty} x^{3 k} (1 - x) d x

= \int_{0}^{1} \frac{x^{3}}{1 + x + x^{2}} d x

= \frac{\sqrt{3} π}{9} - \frac{1}{2}

spelkw

Beginner2022-01-26Added 12 answers

Let's observe the relation with

\sin (2 π \frac{j}{3})

and rewrite this in an elementary way :

\sum_{k = 1}^{\infty} \frac{1}{(3 k + 1) (3 k + 2)} = \sum_{k = 1}^{\infty} \frac{0}{3 k + 0} + \frac{1}{3 k + 1} - \frac{1}{3 k + 2}

= \frac{2}{\sqrt{3}} I (\sum_{j = 1}^{\infty} \frac{e^{2 π i \frac{j}{3}}}{j}) - \frac{1}{1 \cdot 2}

= - \frac{2}{\sqrt{3}} I \ln (1 - e^{2 π \frac{i}{3}}) - \frac{1}{2}

= - \frac{2}{\sqrt{3}} I \ln (\sqrt{3} e^{- π \frac{i}{6}}) - \frac{1}{2}

= - \frac{2}{\sqrt{3}} (- \frac{π}{6}) - \frac{1}{2}

= \frac{\sqrt{3} π}{9} - \frac{1}{2}

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