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Question Number 45527 by Sanjarbek last updated on 14/Oct/18

Commented by Meritguide1234 last updated on 14/Oct/18

Commented by maxmathsup by imad last updated on 14/Oct/18
$let S_n =Σ_(k=1) ^n (k/((2k+1)!!)) we have S_n =(1/2)Σ_(k=1) ^n ((2k+1−1)/((2k+1)!!)) =(1/2)Σ_(k=1) ^n ( (1/((2k−1)!!))−(1/((2k+1)!!)))=(1/2)Σ_(k=1) ^n (v_(k−1) −v_k )withv_k =(1/((2k+1)!!)) S_n =(1/2){v_0 −v_1 +v_1 −v_2 +..+v_(n−1) −v_n )=(1/2){v_0 −v_n } =(1/2){1−(1/((2k+1)!!))} ⇒lim_(n→+∞) S_n =(1/2) we take (2n+1)!!=(2n+1)(2n−1)....3.1$
$l e t S_{n} = \sum_{k = 1}^{n} \frac{k}{(2 k + 1)!!} w e h a v e S_{n} = \frac{1}{2} \sum_{k = 1}^{n} \frac{2 k + 1 - 1}{(2 k + 1)!!}$ $= \frac{1}{2} \sum_{k = 1}^{n} (\frac{1}{(2 k - 1)!!} - \frac{1}{(2 k + 1)!!}) = \frac{1}{2} \sum_{k = 1}^{n} (v_{k - 1} - v_{k}) {w i t h v}_{k} = \frac{1}{(2 k + 1)!!}$ $S_{n} = \frac{1}{2} {v_{0} - v_{1} + v_{1} - v_{2} + . . + v_{n - 1} - v_{n}) = \frac{1}{2} {v_{0} - v_{n}}$ $= \frac{1}{2} {1 - \frac{1}{(2 k + 1)!!}} \Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{1}{2}$ $w e t a k e (2 n + 1)!! = (2 n + 1) (2 n - 1) . . . . 3.1$
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