calculus II

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By definition a series of the form `sum (a_n) ` is absolutely convergent if the series `sum |a_n| ` is also convergent.

To check for the convergence of the series `sum_(n=1)^(+oo) 4/((4n-3)(4n+1)) ` we write the fraction decomposition of it:

`4/((4n-3)(4n+1)) = 1/(4n-3) -1/(4n+1) `

Therefore the series can be written in a simpler form

`sum_(n=1)^(+oo) 4/((4n-3)(4n+1)) =sum_(n=1)^(+oo)(1/(4n-3) -1/(4n+1)) `

Now because in parentheses we have for any `n>=1 ` ,

`1/(4n-3) > 1/(4n+1) ` it results that `sum (a_n) =sum |a_n| `  so that if the first series is convergent it is also absolute


To show that the series ` sum (a_n)` is convergent let us find its sum:

`sum_(n=1)^(+oo) (a_n) =lim_(n->+oo) sum_(i=1)^n (1/(4i-3) -1/(4i+1)) = lim_(n->+oo) [(1-1/5)+(1/5-1/9)+(1/9-1/13)+...+(1/(4n-3)-1/(4n+1))] `

`sum_(n=1)^(+oo) (a_n)=lim_(n->+oo)(1-1/(4n+1)) =1-1/(+oo) =1-0 =1 `

Since the series only have two terms left after cancellation of the other terms it is a telescopic series. Its sum is 1.

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