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What is limit of sum f(1)+f(2)+____+f(n) with n go to inf ? f(x)=1/(x+1)(x+2)

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luvgoj | Student, Undergraduate | (Level 2) Honors

Posted August 23, 2013 at 12:16 PM via web

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What is limit of sum f(1)+f(2)+____+f(n) with n go to inf ?

f(x)=1/(x+1)(x+2)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 23, 2013 at 1:11 PM (Answer #1)

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You need to re-write the equation of the function using partial fraction expansion, such that:

`1/((x + 1)(x + 2)) = a/(x + 1) + b/(x + 2)`

Bringing the terms to a common denominator, yields:

`1 = a(x + 2) + b(x + 1)`

`1 = ax + 2a + bx + b`

Grouping the terms that contain x yields:

`1 = x(a + b) + 2a + b`

Equating the coefficients of equal powers yields:

`{(a + b = 0),(2a + b = 1):} => {(a = -b),(-2b + b = 1):}`

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

`1/((x + 1)(x + 2)) = 1/(x + 1) - 1/(x + 2)`

You need to evaluate the summation `sum_(k=1)^n =`   `f(1) + ... + f(n)` , such that:

`{(f(1) = 1/2 - 1/3),(f(2) = 1/3 - 1/4),(f(3) = 1/4 - 1/5),(.......................),(f(n - 1) = 1/n - 1/(n + 1)),(f(n) = 1/(n + 1) - 1/(n + 2)):} => sum_(k=1)^n f(k) = 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/n - 1/(n + 1) + 1/(n + 1) - 1/(n + 2)`

Reducing duplicate terms yields:

`sum_(k=1)^n f(k) = 1/2 - 1/(n + 2)`

Evaluating the limit of summation yields:

`lim_(n->oo) sum_(k=1)^n f(k) = lim_(n->oo) (1/2 - 1/(n + 2))`

`lim_(n->oo) sum_(k=1)^n f(k) = lim_(n->oo)(1/2) - lim_(n->oo) (1/(n + 2))`

`lim_(n->oo) sum_(k=1)^n f(k) = 1/2 - 0`

`lim_(n->oo) sum_(k=1)^n f(k) = 1/2`

Hence, evaluating the limit of the given summation, using the partial fraction expansion, yields

`lim_(n->oo) sum_(k=1)^n f(k) = 1/2. `

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