help!!!!! series?

Question

can anyone SUM this serie: Sum(when n=1 to infinity) of: (n+1)/(2^n)?
thank you!

Answer 1

This is tricky.

First I'll introduce a Maclaurin series, then use it to sum your series.

The Maclaurin series for 2/(1 - (x/2)) is

2/(1 - (x/2)) = 2∑(x/2)^n,

where the sum is taken from n = 0 to ∞. I may rewrite this as

2/(1 - (x/2)) = 2∑(x^n)/(2^n).

Now take the derivative of both sides:

1/(1 - (x/2))^2 = 2∑(nx^(n - 1))/(2^n),

where the sum is taken from n = 0 to ∞. Now, on the right side, the n = 0 term will be zero, so I may rewrite this as

1/(1 - (x/2))^2 = 2∑(nx^(n - 1))/(2^n)

where the sum is taken from n = 1 to ∞ this time. Now, re-index the sum; start from n = 0 instead of n = 1 and replace all the n's inside the summation with n + 1. This gives

1/(1 - (x/2))^2 = 2∑(n + 1)x^n/(2^(n + 1))

where the sum is taken from n = 0 to ∞. Now bring the 2 that's out front on the inside, cancelling one of the 2's on the bottom on the inside:

1/(1 - (x/2))^2 = ∑(n + 1)x^n/(2^n)

where the sum is taken from n = 0 to ∞. Now plug in x = 1 (which is within the radius of convergence, 2):

1/(1 - 1/2)^2 = ∑(n + 1)/(2^n)

4 = ∑(n + 1)/(2^n).

Since this sum is taken from n = 0 to ∞, then to find your sum, we must subtract off the n = 0 term. The n = 0 term is 1; 4 - 1 = 3.

So your series sums to 3.

Answer 2

I see someone else answered it with McLaurin series. That's the educated way. I forgot how to do that but I did it with what my first year profs called a Cheap Trick, and it's a lot more fun when it works!

first, your series can be written as sum(1 to inf) of n/2^n + sum(1 to inf) of 1/2^n. That's just rearranging the summands.

Now the series sum(1 to inf) 1/2^n famously equals 1. No need to go into detail about that.

Now all that's left is

sum(1 to inf) n/2^n

now for the CHEAP TRICK...

this series looks like

1/2 + 2/4 + 3/8 + 4/16 + 5/32 + 6/64 + ...

can be rearranged like:

= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...
+ 1/4 + 1/8 + 1/32 + 1/64 + ...
+ 1/8 + 1/16 + 1/32 + 1/64 + ...
+ 1/16 + 1/32 + 1/64 + ...

see how that rearranges each of the two fourths, and the "3/8", and the "4/16", etc?

the first line is the same series 1/2^n converging to 1; the second is the same, converging to 1 - 1/2, and so on.

now this sum equals
1
+ (1 - 1/2) ; <---this equals 1/2
+ (1 - 1/2 - 1/4) ;<--this equals 1/4
+ (1 - 1/2 - 1/4 - 1/8) ;<--catching on?
+ ...
= 1 + 1/2 + 1/4 + 1/8 + ...
= 1 + 1 = 2 shazzam!

so if sum(1 to inf) n/2^n = 2, then sum(1 to inf)(n + 1)/2^n = 2 + 1 = 3 shazzam!

I only say shazzam when I have solved something using a cheap trick.

btw I bet this is the simplest answer you get. cheap-trick answers usually are. It all keys off the basic convergent arithmetic series sum(1 to inf)1/2^n = 1, which you can practically prove in your head.

Answer 3

Σ(n+1)/(2^n) = Σn/(2^n) + Σ1/(2^n)
This second series converges to 1. You can show that the first series converges to 2, to make a total sum of 3.

Let S[k] be the partial sum of Σn/(2^n) up to n=k. Then:
S[k] = Σ(n)/(2^n)
(1/2)S[k] = (1/2)Σ(n)/(2^n)
(1/2)S[k] = Σn/2^(n+1)
(1/2)S[k] = Σn=1-to-k n/2^(n+1)
(1/2)S[k] = Σn=2-to-k+1 (n-1)/2^n
(1/2)S[k] = Σn=2-to-k+1 [n/2^n] - Σn=2-to-k+1 [1/2^n]
(1/2)S[k] = Σn=2-to-k+1 [n/2^n] - Σn=1-to-k+1 [1/2^n] + 1/2
(1/2)S[k] = Σn=2-to-k+1 [n/2^n] - Σn=1-to-k [1/2^n] + 1/2 - 1/2^(k+1)
(1/2)S[k] = Σn=2-to-k+1 [n/2^n] - (1 - (1/2)^k) + 1/2 - 1/2^(k+1)
(1/2)S[k] = Σn=2-to-k+1 [n/2^n] - 1/2 + (1/2)^k - 1/2^(k+1)
(1/2)S[k] = Σn=2-to-k+1 [n/2^n] - 1/2 + (1/2)^k (1 - 1/2)
(1/2)S[k] = Σn=2-to-k+1 [n/2^n] - 1/2 + (1/2)^k (1/2)
(1/2)S[k] = Σn=2-to-k+1 [n/2^n] - 1/2 + (1/2)^(k+1)
(1/2)S[k] = Σn=1-to-k+1 [n/2^n] - 1/2 - 1/2 + (1/2)^(k+1)
(1/2)S[k] = Σn=1-to-k+1 [n/2^n] - 1 + (1/2)^(k+1)
(1/2)S[k] = Σn=1-to-k [n/2^n] + (k+1 /2^(k+1)) - 1 + (1/2)^(k+1)
(1/2)S[k] = S[k] + (k+1 /2^(k+1)) - 1 + (1/2)^(k+1)
-(1/2)S[k] = (k+1 / 2^(k+1)) - 1 + (1/2)^(k+1)
S[k] = -(k+1 / 2^k) + 2 + (1/2)^(k)
S[k] = 2 - (k/2^k)

Ugh, I think I messed up some where. But you can find the general proof of the Arithmetico-Geometric Series here:
http://mathforum.org/library/drmath/view/66996.html

Answer 4

it would approach zero

proof01 : divide it by n the equation
gives ( 1+ 1/n ) / ( [2^n]/n )
as n approaches infinity 1/n ->0
also 2^n will grow much larger than n in the denominator
hence (1+0) / inf => 0

proof02: matlab program

Answer 5

sum to infinity when a general term is given-
apply sigma to the term [n+1]/[2^n]
sigma n= n[n+1]/2
use this and simplify to get the answer