calculus II

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You may decide if the sequence is convergent or divergent using both test, the comparison test and integral test, such that:

`ln(n+1)/(n(n+1)) = ln(n+1)/(n^2 + n) < ln(n+1)/(n^2)`

You need to use the integral test to evaluate the convergence of sequence l`n(n+1)/(n^2)` , such that:

`lim_(n->oo) int_1^oo ln(x+1)/(x^2)dx`

You need to use integration by parts to evaluate the integral, such that:

`int ln(x+1)/(x^2)dx = -(ln(x+1))/x + int 1/(x(x+1))`

`ln(x+1) = f(x) => 1/(x+1) = f'(x)`

`x^(-2) = g'(x) => g(x) = -1/x`

You need to use partial fraction decomosition to split the integral into two simpler integrals, such that:

`1/(x(x+1)) = a/x + b/(x+1)`

`1 = x(a+b) + a`

Equating the coefficients of like powers yields:

`{(a + b = 0),(a = 1):} => b = -1`

`1/(x(x+1)) = 1/x - 1/(x+1)`

Integrating both sides, yields:

`int 1/(x(x+1)) dx = int 1/xdx - int 1/(x+1)dx`

`int 1/(x(x+1)) dx = ln(x) - ln(x+1) = ln (x/(x+1))`

`lim_(n->oo) int_1^oo ln(x+1)/(x^2)dx = lim_(n->oo)( -(ln(n+1))/n + ln (n/(n+1)))`

`lim_(n->oo) int_1^oo ln(x+1)/(x^2)dx = lim_(n->oo) ln (n^2/(n+1)^2) = ln lim_(n->oo) (n^2/(n+1)^2) = ln 1 = 0`

Hence, using the integral test, the series `ln(n+1)/(n^2)` converges, thus, by comparison test, the sequence `Sigma_(k=1)^oo (ln(n+1))/(n(n+1))` converges.

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