Proving a Series Using the Digamma Function: A Step-by-Step Guide

In this article, we will explore how to prove the convergence and evaluate the series ( S sum_{n1}^{infty} frac{1}{5n-15n-5n1} ) using the properties of the digamma function. This process involves advanced mathematical concepts, including partial fractions, the digamma function, and trigonometric identities. Let's delve into the step-by-step solution we will explore.

Step 1: Partial Fraction Decomposition

First, we use partial fractions to decompose the given series. The series can be written as:

S frac{1}{2} sum_{n1}^{infty} left( frac{1}{5n-1} - frac{2}{5n} frac{1}{5n 1} right)

Step 2: Applying the Digamma Function

The digamma function, denoted by ( psi(z) ), is the logarithmic derivative of the gamma function. We use the identity for the digamma function for any ( z ) in the complex plane except ( -1, -2, ldots ): [ psi(z 1) - gamma -sum_{k1}^{infty} left( frac{1}{k} - frac{1}{k z} right) ]

Step 3: Simplification Using the Digamma Function

Applying the digamma function to simplify the series, we get:

[ S -frac{1}{10} sum_{n1}^{infty} left[ left( frac{1}{n} - frac{1}{n - frac{1}{5}} right) left( frac{1}{n} - frac{1}{n frac{1}{5}} right) right] -frac{1}{10} left[ left( gamma psi left( -frac{1}{5} right) right) left( gamma psi left( frac{6}{5} right) right) right] ]

Further simplification using the identity ( psi(1-x) - psi(x) pi cot(pi x) ) gives:

[ psi left( frac{4}{5} right) - psi left( frac{1}{5} right) pi cot left( frac{pi}{5} right) ]

Step 4: Evaluating the Digamma Function

Using the Gauss’ Digamma Theorem to evaluate ( psi left( frac{1}{5} right) ), we get:

[ psi left( frac{1}{5} right) -gamma - 5 ln 2 - frac{pi}{2} cot left( frac{pi}{5} right) 2 sum_{n1}^{lfloor frac{5-1}{2} rfloor} cos left( frac{2npi}{5} right) ln sin left( frac{npi}{5} right) ]

Simplifying further, we obtain:

[ S -frac{1}{10} left[ 5 - 2 ln 10 - 4 cos left( frac{2pi}{5} right) ln sin left( frac{pi}{5} right) - 4 cos left( frac{4pi}{5} right) ln sin left( frac{2pi}{5} right) right] ]

Step 5: Simplifying Trigonometric Terms

Further simplification of the trigonometric terms leads to:

[ 4 cos left( frac{2pi}{5} right) ln sin left( frac{pi}{5} right) - 4 cos left( frac{4pi}{5} right) ln sin left( frac{2pi}{5} right) -frac{1}{2} ln left( frac{5 - sqrt{5}}{8} right) left( frac{5 sqrt{5}}{8} right) - frac{sqrt{5}}{2} ln left( frac{3 sqrt{5}}{2} right) ]

And, finally, we find:

[ S -frac{1}{10} left[ 5 - 2 ln 10 frac{1}{2} ln left( frac{16}{5} right) - sqrt{5} ln varphi^2 right] ]

Further rewriting gives:

[ S frac{1}{10} left[ -5 - 2 ln 5 2 ln 2 - frac{1}{2} (4 ln 2 - ln 5) - sqrt{5} ln varphi right] ]

Which simplifies to:

[ S frac{1}{2} left[ -1 - frac{1}{2} ln 5 - frac{1}{sqrt{5}} ln varphi right] frac{ln sqrt{5} varphi^{1/sqrt{5}} - 1}{2} ]

Conclusion

We conclude that the series ( S sum_{n1}^{infty} frac{1}{5n-15n-5n1} frac{ln sqrt{5} varphi^{1/sqrt{5}} - 1}{2} )

References: