what does this series converge to?

Question

1 / [n * (n+2) ] when no goes from 1 to infinity? how do you do it?

Answer 1

Since you mention series, this is a summation, so I'm going to assume that you mean
Σ1 to ∞, 1/[n(n + 2)]

We can use partial fractions to show that

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

Factoring out (1/2) gives us

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

Therefore,

Σ1 to ∞, 1/[n(n + 2)] =

Σ1 to ∞, (1/2)( (1/n) - 1/(n + 2)) =

We can factor out the constant (1/2) from the summation, to obtain

(1/2) Σ1 to ∞ [ (1/n) - 1/(n + 2) ] =

At this point, we expand the summation, to demonstrate that we have terms that cancel each other out.

(1/2) [ (1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + (1/4 - 1/6) +
(1/5 - 1/7) + .... ]

Let's group the positive terms in one set of parentheses, and the negative terms in another set of parentheses. Note that, because all of the terms are negative, we can use a single minus outside of the parentheses to represent all negative terms.

(1/2) [ (1 + 1/2 + 1/3 + 1/4 + ... ) - (1/3 + 1/4 + 1/5 + 1/6 + ... ) ]

Leaving behind 1 + 1/2 and pairing all opposite terms together,

(1/2) [ (1 + 1/2) + (1/3 - 1/3) + (1/4 - 1/4) + (1/5 - 1/5) + ... ]

(1/2) [ 1 + 1/2 + 0 + 0 + 0 + 0 + .... ]

(1/2) (1 + 1/2)
(1/2) (3/2)
3/4

The series converges to 3/4.

Answer 2

it' been a long time, but from what I can remember any fraction that has an infinite as denominator will converge to 0.
if you do it by trials:
if n=1 then 1/(1*(1+2)) = 1/3
n=2 then = 1/4
n=100 then = 1/102
and so on, so it's pretty obvious that it keeps getting smaller and smaller, thus converging to 0

Answer 3

Mercy, what the heck do I not get about this, Puggy? Saying this approaches 3/4 is the same as saying n*(n+2) approaches 4/3 as n approaches infinity - what gives? Am I misreading the question????

[REPOST] aWWWG! I get it, there's supposed to be a Σ sign before the formula, huh? Schit...

Answer 4

Zero.



Let's look at your fraction:

1/ [ n * (n+2)]

the larger that n grows, the larger that [n * (n+2)] will grow

since [n * (n+2)] is a denominator, the larger the denominator grows, the smaller the whole expresion will become.
When n aproaches infinity, then the whole expresion will aproach zero.