SOLUTION: If S=3/2+5/4+7/8+...+(2n+1)/2^n+... where n=1, 2, 3, ..., then find the value of S.

Question 1168321: If S=3/2+5/4+7/8+...+(2n+1)/2^n+... where n=1, 2, 3, ..., then find the value of S.
Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website!

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= =

S = Lim(N --> )
S = Lim(N --> )
S = Lim(N --> )

The terms go to zero, leaving 4+1 = 5

I still needed help (WolframAlpha) to get the first partial sum.

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.

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";}} '

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
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