Calculate limit an=1/(n+3)+1/(n+6)+-----+1/4n?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may factor out `n` to each denominator, such that:

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

Factoring out `1/n,` yields:

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

`a_n = (1/n)*sum_(k=1)^n 1/(1 + 3*(k/n))`

You should notice that a_n represents the Riemann summation associated to the function `f(x) = 1/(1 + 3x)` , hence, you may evaluate the limit of the sequence, such that:

`lim_(n->oo) a_n = int_0^1 f(x)dx`

`int_0^1 f(x)dx = int_0^1 1/(1 + 3x) dx`

You should come up with the following substitution, such that:

`1 + 3x = t => 3dx = dt => dx = (dt)/3`

Changing the limits of integration yields:

`x = 0 => t = 1`

`x = 1 => t = 4`

`int_0^1 1/(1 + 3x) dx = int_1^4 1/t*(dt)/3`

`int_1^4 1/t*(dt)/3 = (1/3) ln t|_1^4`

Using the fundamental theorem of calculus, yields:

`int_1^4 1/t*(dt)/3 = (1/3)(ln 4 - ln 1)`

`int_1^4 1/t*(dt)/3 = (1/3)(ln 4) => int_1^4 1/t*(dt)/3 = ln (root(3) 4)`

Hence, evaluating the limit of the given sequence, using the Riemann summation, yields `lim_(n->oo) a_n = ln (root(3) 4).`

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