find the sum?

Question

∞
Σ 1/n(n+2) = ?
n=1

Answer 1

∞
Σ 1/n(n+2)
n=1
∞
=Σ[1/n -1/(n+2)](1/2)
n=1
=(1/2)[1-1/3+1/2-1/4+1/3-1/5+...+]
= (1/2)(1 + 1/2)
= 3/4
-------------
Reason:
All terms cancel out, except term1 and term3 . Therefore,
1-1/3+1/2-1/4+1/3t-1/5+...+
= 1+1/2

Answer 2

Re: One of the previous answers. Complex contour integration to find a sum? No. Contour integration is more something to find the integral of a function over the entire real line.

To set this up, first we're going to do the partial fractions thing. Sorry about the clunky look of what follows, but Yahoo's insistence on programming this thing to eliminate whitespace forced me to use a less easily readable layout than I'd normally use.

a /n + b/(n+2) = 1

a (n + 2) + bn = 1

(a + b) n + 2a = 1

so: a+b=0, 2a = 1
a+b=0 a=1/2
implying: b= -1/2

thus:

(1/2)/n + (-1/2)/(n+2) = 1

∞
Σ 1/n(n+2) =
n=1

∞
Σ (((1/2) /n) - ((1/2)/(n+2)) =
n=1


∞
( Σ (1/n - 1/(n+2))) (1/2) =
n=1

∞
(Σ 1/n -
n=1

∞
Σ 1/(n+2) ) (1/2) =


∞
(Σ 1/n -
n=1

∞
Σ 1/m) (1/2) =
m=3

(substituting m for n+2 in the
second sum)


∞
(Σ 1/m -
m=1

∞
Σ 1/m) (1/2) =
m=3


(substituting m for n in the first
sum)



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

= (3/2) (1/2) = 3/4


since the second sum is just the first sum, with the first two terms removed.

Answer 3

Σ 1/n(n+2) = Σ{1/n - 1/(n+2)}/2
2 x sum = 1/1 - 1/3 + 1/2 - 1/4 + 1/3 + 1/5 + 1/4 - 1/6 + ...
All terms cancel except 1/1 and 1/2,. So
2 x sum = 1 + 1/2.
So your sum = 3

Th

Answer 4

I suppose you mean 1/{ n(n+2) }

( if not then the sum has no limit )

I know that :

∞
Σ 1/n^2 = pi^2/6
n=1



Am i helpfull ?

1/(n+2)^2 < 1/{ n(n+2) } < 1/n^2

so its between the left sum and the right sum

pi^2/6 - 1 - 1/4 < your sum < pi^2/6

i think you should do this with complex contour integral around the singularities z=0 and z=-2 w

welll i dont know anymore

Answer 5

3

Answer 6

It is curious indeed how simple and elegant methods can be stifled by an unnecessary application of complex analysis.

Back to the problem, simply observe that this sum is indeed
lim N
N->∞ Σ 1/n(n+2)
n=1
Now break the general term into partial fractions and notice how terms cancel out leaving just a few as they were.
Now find the limit as N->∞

Voila!
Hope this helps!

Answer 7

3/4

write 1/n(n+2) as 1/2*(1/n-1/(n+2)) and then write out the first few terms of each of these sums. You'll see that all the terms cancel out apart from the first 2 terms from the first sum, which are 1/2*1/1+1/2*1/2=3/4.

Answer 8

what problem number is this..