let-S-n-k-1-n-1-k-k-1-calculate-S-n-interms-of-n-2-find-lim-n-S-n-

Question Number 69374 by mathmax by abdo last updated on 22/Sep/19

l e t S_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{k}

1) c a l c u l a t e S_{n} i n t e r m s o f n

2) f i n d {l i m}_{n \to + \infty} S_{n}

Commented by mathmax by abdo last updated on 15/Oct/19

1) S_{n} = W (- 1) w i t h w (x) = \sum_{k = 1}^{n} \frac{x^{k}}{k}

w^{^{'}} (x) = \sum_{k = 1}^{n} x^{k - 1} = \sum_{k = 0}^{n - 1} x^{k} = \frac{x^{n} - 1}{x - 1} (w e s u p p o s e x \neq 1) \Rightarrow

w (x) = \int_{0}^{x} \frac{t^{n} - 1}{t - 1} d t + c w i t h c = w (0) = 0 \Rightarrow w (x) = \int_{0}^{x} \frac{t^{n}}{t - 1} d t - \int_{0}^{x} \frac{d t}{t - 1}

= \int_{0}^{x} \frac{t^{n}}{t - 1} d t - l n ∣ x - 1 ∣\Rightarrow S_{n} = w (- 1) = \int_{0}^{- 1} \frac{t^{n}}{t - 1} d t - l n (2)

t = - u g i v e \int_{0}^{- 1} \frac{t^{n}}{t - 1} d t = \int_{0}^{1} \frac{{(- u)}^{n}}{- u - 1} (- d u) = {(- 1)}^{n} \int_{0}^{1} \frac{u^{n}}{1 + u} d u \Rightarrow

S_{n} = {(- 1)}^{n} \int_{0}^{1} \frac{u^{n}}{1 + u} d u - l n (2)

2) w e h a v e ∣ S_{n} + l n (2) ∣ = \int_{0}^{1} \frac{u^{n}}{1 + u} d u ⩽ \int_{0}^{1} u^{n} d u = \frac{1}{n + 1} \Rightarrow

{l i m}_{n \to + \infty} ∣ S_{n} + l n (2) ∣= 0 \Rightarrow {l i m}_{n \to + \infty} S_{n} = - l n (2) .