Question

How about another sequence... (equation to follow)

Answer 1

Use \[\sum_{n=1}^{\infty} (n-1)/(2^{n+1})=1/2 \]
to calculate
\[\sum_{n=1}^{\infty}n/2^n\]

I know the answer is 2, but not sure how to calculate.

Thanks,

Answer 2

Sorry lol hav no idea.

Answer 3

If the first equation is (A) and second equation is (B), I already know that (A) = 1/2. Since (A) = (B) - [(n+1)/(2^k+1)] I can get through some algebra... But I don't htink taht's helping. I'll come back to it later I guess!

Answer 4

n * 2^(-n)

the sum of a product is the product of the sums; right? i never remember that correctly

Answer 5

I thought that the sum would be infinite, since you can always group the resulting fractions to make 1/2 (or 1/4 or whatever). So I'm confused. :)

Answer 6

you mean a telescoping sum? maybe

Answer 7

telesco[ing would have to have an alternating +- tho ... so that ideas off

Answer 8

Right. THat's how I got to (A) = 1/2.

Answer 9

Crap I have to go to a meeting. Thanks for thinking about this and responding. I'll check it out when I get back.

Answer 10

1/2

1/2 + 2/4 = (2+2)/4

(2+2)/4 + 3/8 = (4+4+3)/8

(4+4+3)/8 + 4/16 = (8+8+6+4)/16

sooo

(16+16+12+8+5)/32

double the top, double the bottom, add the next "n" to the top

(32+32+24+16+10+6)/64

\[a_{n+1}=\frac{2(a_n)+(n+1)}{2^{n+1}}\]
looks like something that might be a hassle to iron out into partial sums

Answer 11

\[\sum_{n=1}^{\infty} (n-1)/(2^{n+1})=1/2\]

\[\sum_{n=1}^{\infty} n/2^{n+1} -\sum_{n=1}^{\infty} 1/2^{n+1} =1/2\]
\[\sum_{n=1}^{\infty} 1/2^{n+1} =1/2\hspace{1cm}\text{ geometric sum}\]

\[\Rightarrow\sum_{n=1}^{\infty} n/2^{n+1} =1\]
\[\Rightarrow\frac{1}{2}\sum_{n=1}^{\infty} n/2^{n} =1\]
\[\Rightarrow\sum_{n=1}^{\infty} n/2^{n} =2\]

Answer 12

i can never seem to get those things to stick in my head :)