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Calculate limit of series an=1/(n+2)+1/(n+4)+----+1/3n?

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minlux | (Level 2) Honors

Posted September 12, 2013 at 2:50 PM via web

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Calculate limit of series an=1/(n+2)+1/(n+4)+----+1/3n?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 12, 2013 at 3:01 PM (Answer #1)

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You need to re-write the given seqeunce, such that:

`a_n = 1/(n + 2) + 1/(n + 2*2) + 1/(n + 2*3) + ... + 1/(n + 2*n)`

Factoring out `1/n` yields:

`a_n =(1/n)*(1/(1 + 2*1/n) + 1/(1 + 2*2/n) + ... + 1/(1 + 2*n/n))`

`a_n =(1/n)*sum_(k=1)^n 1/(1 + 2*(k/n))`

You need to notice that the summation is a Riemann summation associated to the function `f(x) = 1/(1 + 2x)` , hence, you may evaluate the limit of the given sequence, such that:

`lim_(n->oo) a_n = int_0^1 1/(1 + 2x) dx`

You should use integration by substitution, such that:

`1 + 2x = y => 2dx = dy => dx = (dy)/2`

`x = 0 => y = 1`

`x = 1 => y = 3`

`int_0^1 1/(1 + 2x) dx = int_1^3 1/y * (dy)/2`

`int_1^3 1/y * (dy)/2 = (1/2) ln y|_1^3`

By fundamental theorem of calculus, yields:

`int_1^3 1/y * (dy)/2 = (1/2) (ln 3 - ln 1)`

`int_1^3 1/y * (dy)/2 = (1/2) (ln 3)`

Using the power property of logarithms yields:

`int_1^3 1/y * (dy)/2 = ln sqrt 3`

Hence, evaluating the limit of the given sequence, using Riemann summation, yields `lim_(n->oo) a_n = ln sqrt 3. `

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