# find the limit of the sum (f(1)+f(2)+.....+f(n))^(n^2), if n approaches to infinite?

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### 2 Answers

We'll try to express each term of the sum as a difference of 2 elemetary fractions.

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

(2x+1)/x^2*(x+1)^2 = A/x + B/x^2 + C/(x+1) + D/(x+1)^2

We'll multiply each term from the right side by LCD = x^2*(x+1)^2 and we'll get:

2x + 1 = Ax(x+1)^2 + B(x+1)^2 + Cx^2(x+1) + Dx^2

2x + 1 = Ax^3 + 2Ax^2 + Ax + Bx^2 + 2Bx + B + Cx^3 + Cx^2 + Dx^2

2x + 1 = x^3(A + C) + x^2(2A + B + C + D) + x(A + 2B) + B

Comparing both sides, we'll get:

A + C = 0

2A + B + C + D = 0

A + 2B = 2

B = 1 => A + 2 = 2 => A=C=0

D = -1

(2x+1)/x^2*(x+1)^2 = 1/x^2 - 1/(x+1)^2

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

For x = 1 => f(1) = 1/1^2 - 1/(1+1)^2 = 1 - 1/2^2

For x = 2 => f(2) = 1/2^2 - 1/(3)^2

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For x = n => f(n) = 1/n^2 - 1/(n+1)^2

f(1) + f(2) + ... + f(n) = 1 - 1/2^2 + 1/2^2 - 1/3^2 + ... + 1/n^2 - 1/(n+1)^2

We'll eliminate like terms:

f(1) + f(2) + ... + f(n) = 1 - 1/(n+1)^2

We'll raise to n^2 both sides:

[f(1) + f(2) + ... + f(n)]^(n^2) = [1 - 1/(n+1)^2]^(n^2)

We'll evaluate the limit:

lim [f(1) + f(2) + ... + f(n)]^(n^2) = lim [1 + 1/-(n+1)^2]^(n^2)

We'll create remarcable limit "e":

lim [1 + 1/-(n+1)^2]^(n^2) = e^lim -n^2/(n+1)^2

lim -n^2/(n+1)^2 = lim -n^2/(n^2 + 2n + 1)

lim -n^2/(n^2 + 2n + 1) = 1

lim [f(1) + f(2) + ... + f(n)]^(n^2) = e^-1

lim [f(1) + f(2) + ... + f(n)]^(n^2) = 1/e

**The result of the limit of the given sum, if n approaches to infinite, is: lim [f(1) + f(2) + ... + f(n)]^(n^2) = 1/e.**

the function is f(x)=(2x+1)/x^2*(x+1)^2