Check the solution to "How to prove:∑n=1∞(14n+1−14n)=18(π−8+6ln⁡2)"

Ingrid Senior

Answered question

2022-02-24

How to prove:

\sum_{n = 1}^{\infty} (\frac{1}{4 n + 1} - \frac{1}{4 n}) = \frac{1}{8} (π - 8 + 6 \ln 2)

Warenberg56i

Beginner2022-02-25Added 8 answers

Another approach is to use the Anti-difference concept.
Since the digamma function is defined as

ψ (z) = \frac{d}{d z} \ln Γ (z)

its functional equation is

△ ψ (z) = ψ (z + 1) - ψ (z) = \frac{d}{d z} \ln (z Γ (z)) - \frac{d}{d z} \ln (Γ (z)) = \frac{d}{d z} \ln z = \frac{1}{z}

It follows that

\sum_{n = 1}^{N} \frac{1}{n + a} = \sum_{n = 1}^{N} ψ (n + a + 1) - ψ (n + a) = ψ (N + a + 1) - ψ (1 + a)

and therefore

\sum_{n = 1}^{N} \frac{1}{4 n + 1} - \frac{1}{4 n} = \frac{1}{4} \sum_{n = 1}^{N} \frac{1}{n + \frac{1}{4}} - \frac{1}{n}

= \frac{1}{4} (ψ (N + \frac{5}{4}) - ψ (\frac{5}{4}) - ψ (N + 1) + ψ (1))

Since

ψ (z)

is holomorphic for

0 < R (z)

, then

ψ (N + \frac{5}{4}) - ψ (N + 1) = ψ^{(1)} (N + 1) \frac{1}{4} + \frac{ψ^{(2)} (N + 1)}{2!} {(\frac{1}{4})}^{2} + \dots

and since

lim_{N \to \infty} ψ^{(k)} (N + 1) = 0 ∣ 1 \leq k

Therefore

lim_{N \to \infty} \sum_{n = 1}^{N} \frac{1}{4 N + 1} - \frac{1}{4 n} = \frac{1}{4} (ψ (1) - ψ (\frac{5}{4})) = \frac{3}{4} \ln 2 + \frac{π}{8} - 1

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