Question

find the sum from n=1 to infinity of (n/2^n)

Answer 1

http://www.wolframalpha.com/input/?i=sum%20from%20n%3D1%20to%20infinity%20of%20(n%2F2%5En)&t=crmtb01&f=rc

I wish I could help more than that, but I honestly don't know.

Answer 2

i tried it but the finial answer is 2 but im not sure if its right?

Answer 3

Hmmmm, I wonder...

Answer 4

Sometimes binomial theorem helps, but I don't think it would in this case.

Answer 5

Well it definitely converges.....but I'm not sure you can get an ACTUAL answer for a sum.

Answer 6

i don't know which method to use. I understand how to solve
sum from n=1 to infinity of (1/2^n). but because there is an n in the numerator i couldn't solve it

Answer 7

I don't think you can get an actual sum, I think you just prove it converges.

Answer 8

the problem is the question says find the sum

Answer 9

Unless it's a geo-series or a telescoping series, I have no idea if there is a way to find the exact sum. Obviously there must be if it's asking for one, but that's new to me.

Answer 10

I proved that its convergent using the ratio test but i dont know how to find the sum

Answer 11

Hmmm, I wonder what happens if you substitute \(n\) with \(n+1\) and try to simplify...

Answer 12

Yeah, I did the same.

Answer 13

Ratio test to prove convergence and then I didn't know what more to do, lol

Answer 14

\[
\frac n {2^n}
\]Put in \(n+1\):\[
\frac {n+1} {2^{n+1}} = \frac1 2\left(\frac {n+1}{2^n}\right) = \frac 1 2 \left( \frac n {2^n} + \frac 1 {2^n}\right)
\]

Answer 15

I dunno if this helps but it's something to think about.

Answer 16

\[
a_n = \frac 1 2 \left(a_{n-1} + \frac 1 {2^n} \right)
\]

Answer 17

Old trick

1) \(\dfrac{1}{2} + \dfrac{2}{2^{2}} + \dfrac{3}{2^{3}} + \dfrac{4}{2^{4}} + ... = S\)

Multiply by 1/2 (Doesn't seem like it will get us anywhere, does it?)

2) \(\dfrac{1}{2^{2}} + \dfrac{2}{2^{3}} + \dfrac{3}{2^{4}} + \dfrac{4}{2^{5}} + ... = \dfrac{S}{2}\)

Subtract 2 from 1

\(\dfrac{1}{2} + \dfrac{1}{2^{2}} + \dfrac{1}{2^{3}} + \dfrac{1}{2^{4}} + ... = \dfrac{S}{2}\)

Now, the left hand side is just a geometric series.

\(\dfrac{\dfrac{1}{2}}{1 - \dfrac{1}{2}} = 1 = \dfrac{S}{2} \implies S = 2\)

Answer 18

@tkhunny The step in which you subtracted 2 from 1, what exactly did you do?

Answer 19

Yeah, thats what Im looking at.

Answer 20

It sort of telescopes. Each denominator matches up nicely between 1) and 2), with the exception of the first term.

Answer 21

this is what i did so far
\[\sum_{1}^{\infty} (\frac{ n }{ 2^n }) = \sum_{1}^{\infty} (\frac{ n-1+1 }{ 2^n }) = \sum_{1}^{\infty} (\frac{ n-1 }{ 2^n } +\frac{ 1 }{ 2^n } )= \sum_{1}^{\infty} (\frac{ n-1 }{ 2^n }) + \sum_{1}^{\infty} (\frac{ 1 }{ 2^n } )\]
and I know that
\[\sum_{1}^{\infty} (\frac{ 1 }{ 2^n }) = 1\]
which means that
\[\sum_{1}^{\infty} (\frac{ n-1 }{ 2^n }) =1\] but how to prove that
\[\sum_{1}^{\infty} (\frac{ n-1 }{ 2^n })\] is really equals 1

Answer 22

@Brittny You are basically circling the problem. To prove that the n-1/2^n equals one, we really need to prove that n/2^n equals 2 which brings us back to where we started.

Answer 23

http://www.wolframalpha.com/input/?i=sum%20from%20n%3D1%20to%20infinity%20of%20%28n%2F2^n%29&t=crmtb01&f=rc

Answer 24

@tkhunny I see what you did now. I'm trying to generate another solution though. Let's hope it works out =]

Btw that's the kind of trick one uses to prove the sum formula for an arithmetic sequence; but it doesn't always come to mind ;]

Answer 25

Then you do the subtraction.

Answer 26

Standard fare for linearly-increasing benefits from the world of actuaries. :-)

Answer 27

@tkhunny u becoming an actuary?

Answer 28

@tkhunny why did u multiply by 1/2 it seems that any number could work and i would get an infinite number of solutions

Answer 29

@genius12 ahaha! No, already been doing that for 30 years!

@Brittny 1/2 is the common ratio for what I knew ahead of time would be a geometric series.. Always the common ratio.

Answer 30

@tkhunny WAAAAAAAAAAAAAAAAAT?@!?#!@?#@!#?!@#@?!#?# SRS BRO? lol i thought u were my age =.=

Answer 31

That picture of me singing opera? It's more than 15 years old!

Answer 32

Sorry, wrong opera. That one is only about 5 years old. My bad.

Answer 33

to find the sum of the seires.
we took in school its done by two ways , making the seires as geo seires or telescoping series

Answer 34

i don't know if that helps

Answer 35

Wish there was another example for that trick.

Answer 36

I think this is how you would generate a formula for these type of series. First start with a finite geometric series:\[1+x+x^2+\cdots +x^n=\frac{x^{n+1}-1}{x-1}.\]Then take the derivative of both sides:\[1+2x+3x^2+\cdots +nx^{n-1}=\frac{(n+1)x^n(x-1)-(x^{n+1}-1)}{(x-1)^2}\]\[=\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}.\]Multiply both sides by x:\[x+2x^2+3x^3+\cdots +nx^n=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}\]In summation form this is saying:\[\sum_{k=1}^{n}kx^k=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}\]Now take n to infinity. If |x|<1, it converges, since both:\[nx^{n+2},(n+1)x^{n+1}\]go to zero. If |x|<1, we end up with:\[\sum_{k=0}^{\infty}kx^k=\frac{x}{(x-1)^2}.\]If you let x = 1/2, then you get 2 as an answer. Not entirely sure if this is correct, I'll try it out with some other series of this form.

Answer 37

It works for:\[\sum_{n=1}^\infty \frac{n}{3^n}=\frac{3}{4}\] http://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+infinity+of+%28n%2F3%5En%29

and:\[\sum_{n=1}^\infty \frac{n}{5^n}=\frac{5}{16}\] http://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+infinity+of+%28n%2F5%5En%29

Answer 38

thanks a lot

Answer 39

@Brittny "...find the sum of the seires. We took in school its done by two ways , making the series as geo series or telescoping series "

This is good. In this case, in my demonstration, the original series was made to "telescope" into a geometric series. The is a little different from telescoping that annihilates everything, but it is still the familiar "telescoping". You need both concepts to solve this one.