if-k-1-n-u-k-n-2-n-3-determine-lim-n-k-1-n-1-u-k-

Question Number 108705 by mathmax by abdo last updated on 18/Aug/20

if \sum_{k = 1}^{n} u_{k} = n (2^{n} + 3) determine \lim_{n \to + \infty} \sum_{k = 1}^{n} \frac{1}{u_{k}}

Answered by mathmax by abdo last updated on 21/Aug/20

we have u_{1} + u_{2} + \dots . + u_{n} = n (2^{n} + 3)

u_{1} + u_{2} + \dots . + u_{n} + u_{n + 1} = (n + 1) (2^{n + 1} + 3) \Rightarrow

u_{n + 1} = (n + 1) 2^{n + 1} + 3 n + 3 - {n 2}^{n} - 3 n

= 2 n 2^{n} + 2^{n + 1} - {n 2}^{n} + 3 = n 2^{n} + 2 . 2^{n} + 3 = (n + 2) 2^{n} + 3 \Rightarrow

u_{n} = (n + 1) 2^{n - 1} + 3 \Rightarrow v_{n} = \sum_{k = 1}^{n} \frac{1}{u_{k}} = \sum_{k = 1}^{n} \frac{1}{(k + 1) 2^{k - 1} + 3}

1 ⩽ k ⩽ n \Rightarrow 2 ⩽ k + 1 ⩽ n + 1 and 1 ⩽ 2^{k - 1} ⩽ 2^{n - 1} \Rightarrow 2 ⩽ (k + 1) 2^{k - 1} ⩽ n + 1 + 2^{n - 1} \Rightarrow

5 ⩽) k + 1) 2^{k - 1} + 3 ⩽ n + 4 + 2^{n - 1} \Rightarrow \frac{1}{n + 4 + 2^{n - 1}} ⩽ \frac{1}{(\dots)} ⩽ \frac{1}{5}

\frac{1}{n + 4 + 2^{n - 1}} = \frac{1}{2^{n - 1} {\frac{n + 4}{2^{n - 1}} + 1}} \sim \frac{1}{2^{n - 1}} (1 - \frac{n + 4}{2^{n - 1}}) = \frac{1}{2^{n - 1}} - \frac{n + 4}{4^{n - 1}}

\Rightarrow Σ \frac{1}{n + 4 + 2^{n - 1}} converges \Rightarrow Σ v_{n} cv rest to find limit \dots .

be continued \dots .