Question 1168321
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Re-visit... 

 {{{S[N] = sum((2k+1)/2^k, 1, N) }}} = {{{ sum(((2k)/2^k),1,N) + sum((1/(2^k)), 1, N)}}} =  {{{ (2/2^N)(-N+2^(N+1)-2) + (1-2^(-N)) }}}<br>
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S = Lim(N --> {{{infinity}}}) {{{ (2/2^N)(-N+2^(N+1)-2) + (1-2^(-N))) }}}
S = Lim(N --> {{{infinity}}}) {{{ ((-2N/2^N)+(2^(N+2)/2^N)-(2/2^N)) + (1-2^(-N))) }}}
S = Lim(N --> {{{infinity}}}) {{{ (red(-2N/2^N)+(2^(N+2)/2^N)-red(2/2^N)) + (1-red(2^(-N)))) }}}
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The {{{red(red)}}} terms go to zero,  leaving 4+1 = 5
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I still needed help (WolframAlpha) to get the first partial sum.  

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While I can't derive the answer, here is a little script that sums the first 1000 terms.  I verified this answer on Wolfram Alpha.
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perl -e '$n=1000;$s=0;for($i=1;$i<=$n;$i++) {$s+=(2*$i+1)/(2**$i); if ($i<=20 || !($i%100)) { print "$i terms: sum is $s\n";}} '
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1 terms: sum is 1.5
2 terms: sum is 2.75
3 terms: sum is 3.625
4 terms: sum is 4.1875
5 terms: sum is 4.53125
6 terms: sum is 4.734375
7 terms: sum is 4.8515625
8 terms: sum is 4.91796875
9 terms: sum is 4.955078125
10 terms: sum is 4.9755859375
11 terms: sum is 4.98681640625
12 terms: sum is 4.992919921875
13 terms: sum is 4.9962158203125
14 terms: sum is 4.99798583984375
15 terms: sum is 4.99893188476562
16 terms: sum is 4.99943542480469
17 terms: sum is 4.99970245361328
18 terms: sum is 4.99984359741211
19 terms: sum is 4.99991798400879
20 terms: sum is 4.99995708465576
100 terms: sum is 5
200 terms: sum is 5
300 terms: sum is 5
400 terms: sum is 5
500 terms: sum is 5
600 terms: sum is 5
700 terms: sum is 5
800 terms: sum is 5
900 terms: sum is 5
1000 terms: sum is 5