Nice-Calculus-Evaluate-n-1-1-

Question Number 138823 by mnjuly1970 last updated on 18/Apr/21

\dots \dots N i c e \dots C a l c u l u s \dots \dots

E v a l u a t e :: ϕ = \underset{n = 1}{\sum^{\infty}} \frac{1}{} = ?

Answered by qaz last updated on 18/Apr/21

ϕ = \underset{n = 1}{\sum^{\infty}} \frac{1}{(\begin{matrix} n + 2 \\ 3 \end{matrix})} = \underset{n = 1}{\sum^{\infty}} \frac{6}{(n + 2) (n + 1) n}

= 6 \int_{0}^{1} d x \int_{0}^{x} d x \int_{0}^{x} \underset{n = 1}{\sum^{\infty}} x^{n - 1} d x

= 6 \int_{0}^{1} d x \int_{0}^{x} d x \int_{0}^{x} \frac{d x}{1 - x}

= - 6 \int_{0}^{1} d x \int_{0}^{x} l n (1 - x) d x

= 6 \int_{0}^{1} x l n x + 1 - x d x

= \frac{9}{2}

Commented by mnjuly1970 last updated on 18/Apr/21

g r a t e f u l \dots