find the sum of the convergent series: n=1 {sum} to infinity of [ 2 / (n(n+1)) ]?

Answer 1

Whenever you see a problem like that, think partial fractions.

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

∑ = ∑ (2/n) - ∑ 2/(n+1)
= [2 + 2/2 + 2/3 + 2/4 + ...] - [2/2 + 2/3 + 2/4 + ...]
So you can see that every term cancels off except 2

∑ = 2

Answer 2

2

Answer 3

Since 1/(n(n+1))= 1/n - 1/(n+1) the N-partial sum is 1 - 1/(N+1) which tends to 1 as N tends to infinity. Multiply by 2 and you get the answer: 2.

Answer 4

The sequences of the partial sums are drawing close closer and closer to 0. Edit: I transformed the project and took the partial fractions strategies-set. in case you probably did mean from 0 to infinity as you published from before then the effect for that vital ought to come out as " (a million/9)ln(3n - a million) - (a million/9)ln(3n + 2)" = (a million/9)ln(3n - a million)/(3n + 2)) changing infinity by making use of "N" and protecting the decrease decrease 0. the top decrease is going to 0 as N --> infinity. It finally ends up with "(-a million/9)ln(-a million/2)" that may not the sum of the countless sequence although. The vital computes the area below the curve. This fee could desire to the two be larger or below the actually sequence