calculate-n-1-1-n-n-by-using-digamma-

Question Number 174118 by Mathspace last updated on 25/Jul/22

c a l c u l a t e \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} b y u s i n g

ψ (d i g a m m a)

Answered by mnjuly1970 last updated on 25/Jul/22

\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n} = \underset{n = 1}{\sum^{\infty}} \frac{1}{2 n} - \underset{n = 1}{\sum^{\infty}} \frac{1}{2 n - 1}

= \frac{1}{2} {\underset{n = 1}{\sum^{\infty}} \frac{1}{n} - \underset{n = 1}{\sum^{\infty}} \frac{1}{n - \frac{1}{2}}}

= \frac{1}{2} {l i m}_{n \to \infty} {γ - γ + \underset{k = 1}{\sum^{n}} (\frac{1}{k} - \frac{1}{k - \frac{1}{2}})}

= \frac{γ}{2} + \frac{1}{2} ψ (\frac{1}{2}) = \frac{γ}{2} + \frac{1}{2} (- γ - 2 l n (2))

= - l n (2)