Resolved: "How to prove that ∑n=1∞(n−1)!n∏i=1n(a+i)=∑k=1∞1(a+k)2 for a≻1?"

Breanna Mcclure

Answered question

2022-04-06

How to prove that

\sum_{n = 1}^{\infty} \frac{(n - 1)!}{n \prod_{i = 1}^{n} (a + i)} = \sum_{k = 1}^{\infty} \frac{1}{{(a + k)}^{2}}

for

a ≻ 1

Answer & Explanation

firenzesunzc65

Beginner2022-04-07Added 16 answers

This uses a reliable trick with the Beta function. I say reliable because you can use the beta function and switching of the integral and sum to solve many series very quickly.
First notice that

\prod_{i = 1}^{n} (a + i) = \frac{Γ (n + a + 1)}{Γ (a + 1)}

Then

\frac{(n - 1)!}{\prod_{i = 1}^{n} (a + i)} = \frac{Γ (n) Γ (a + 1)}{Γ (n + a + 1)} = B (n, a + 1) = \int_{0}^{1} {(1 - x)}^{n - 1} x^{a} d x

Hence, upon switching the order we have that

\sum_{n = 1}^{\infty} \frac{(n - 1)!}{n \prod_{i = 1}^{n} (a + i)} = \int_{0}^{1} x^{a} (\sum_{n = 1}^{\infty} \frac{{(1 - x)}^{n - 1}}{n}) d x

Recognizing the power series, this is

\int_{0}^{1} x^{a} \frac{- \log x}{1 - x} d x

Now, expand the power series for

\frac{1}{1 - x}

to get

\sum_{m = 0}^{\infty} - \int_{0}^{1} x^{a + m} \log x d x

It is not difficult to see that

- \int_{0}^{1} x^{a + m} \log x d x = \frac{1}{{(a + m + 1)}^{2}}

so we conclude that

\sum_{n = 1}^{\infty} \frac{(n - 1)!}{n \prod_{i = 1}^{n} (a + i)} = \sum_{m = 1}^{\infty} \frac{1}{{(a + m)}^{2}}

Hope that helps,

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