We have an answer to "Evaluate ∑n,k1(nk) where the summation ranges over all..."

siutulysr5

Answered question

2022-01-23

Evaluate

\sum_{n, k} \frac{1}{(\begin{matrix} n \\ k \end{matrix})}

where the summation ranges over all positive integers n, k with

1 < k < n - 1

Frauental91

Beginner2022-01-24Added 15 answers

Let

l = n - k

, we have

\sum_{n = 4}^{\infty} \sum_{k = 2}^{n - 2} {(\begin{matrix} n \\ k \end{matrix})}^{- 1} = \sum_{k = 2}^{\infty} \sum_{n = k + 2}^{\infty} {(\begin{matrix} n \\ k \end{matrix})}^{- 1}

= \sum_{k = 2}^{\infty} \sum_{l = 2}^{\infty} {(\begin{matrix} k + l \\ k \end{matrix})}^{- 1} = \sum_{k = 2}^{\infty} \sum_{l = 2}^{\infty} \frac{k! l!}{(k + l)!}

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

= \sum_{k = 2}^{\infty} \sum_{l = 2}^{\infty} (k + l + 1) \int_{0}^{1} t^{k} {(1 - t)}^{l} d t

= \int_{0}^{1} \sum_{k = 2}^{\infty} \sum_{l = 2}^{\infty} [(k + l + 1) t^{k} {(1 - t)}^{l}] d t

Notice when s and t are independent, we have

\sum_{k = 2}^{\infty} \sum_{l = 2}^{\infty} (k + l + 1) t^{k} s^{l} = \sum_{k = 2}^{\infty} \sum_{l = 2}^{\infty} (t \frac{d}{d t} + s \frac{d}{d s} + 1) t^{k} s^{l}

= (t \frac{d}{d t} + s \frac{d}{d s} + 1) \frac{t^{2} s^{2}}{(1 - t) (1 - s)}

= \frac{s^{2} t^{2} (5 - 4 (s + t) + 3 s t)}{{(1 - s)}^{}}

Aaron Hughes Beginner2022-01-25Added 13 answers

Partial Fractions and Telescoping Sums change order of summation

\sum_{n = 4}^{\infty} \sum_{k = 2}^{n - 2} \frac{1}{(\begin{matrix} n \\ k \end{matrix})} = \sum_{k = 2}^{\infty} \sum_{n = k + 2}^{\infty} \frac{1}{(\begin{matrix} n \\ k \end{matrix})}

= \sum_{k = 2}^{\infty} \frac{k}{k - 1} \sum_{n = k + 2}^{\infty} (\frac{1}{(\begin{matrix} n - 1 \\ k - 1 \end{matrix})} - \frac{1}{(\begin{matrix} n \\ k - 1 \end{matrix})})

Telescoping sum

= \sum_{k = 2}^{\infty} \frac{k}{k - 1} \frac{1}{(\begin{matrix} k + 1 \\ k - 1 \end{matrix})}

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

Partial fractions

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

Telescoping sum

= 1 + \frac{1}{2}

Simplify

= \frac{3}{2}

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