FInd the integral of the infinite series of (2x)^n / n^2
I got (2x)^(n+1) / 2n^2(n+1)
is this right
∫ from 0 to ∞ ∑ (2x)^n / n^2
almost. You forgot the chain rule:
∫(2x)^n/n^2 = (2x)^(n+1)/(n^2 * (n+1)*2)
That's literally what i had
You are correct, but we were both wrong. It should have been
(2x)^(n+1)/(n^2 * (n+1)) * 2
why that way
To find the integral of the infinite series ∫ from 0 to ∞ ∑ (2x)^n / n^2, you can use the technique of power series expansion.
First, let's consider the power series expansion of (1-x)^(-2):
(1-x)^(-2) = 1 + 2x + 3x^2 + 4x^3 + ...
Now, we differentiate both sides of the equation with respect to x:
d/dx [(1-x)^(-2)] = d/dx [1 + 2x + 3x^2 + 4x^3 + ...]
Differentiating each term on the right-hand side with respect to x:
2(1-x)^(-3) = 2 + 6x + 12x^2 + 20x^3 + ...
Now, let's integrate both sides of the equation with respect to x:
∫ [2(1-x)^(-3)] dx = ∫ [2 + 6x + 12x^2 + 20x^3 + ...] dx
Integrating each term on the right-hand side:
-2(1-x)^(-2) = 2x + 3x^2 + 4x^3 + ...
Now, let's substitute x = 2x into the equation:
-2(1 - 2x)^(-2) = 2(2x) + 3(2x)^2 + 4(2x)^3 + ...
Simplifying the equation:
-2(1 - 2x)^(-2) = 4x + 12x^2 + 32x^3 + ...
Finally, let's integrate both sides of the equation with respect to x:
∫ [-2(1 - 2x)^(-2)] dx = ∫ [4x + 12x^2 + 32x^3 + ...] dx
Integrating each term on the right-hand side:
- (1 - 2x)^(-1) = 2x^2 + 4x^3 + 8x^4 + ...
Therefore, the integral of the given series is - (1 - 2x)^(-1) + C, where C is a constant of integration.
So, your answer of (2x)^(n+1) / (2n^2(n+1)) is incorrect. The correct integral of the given series is - (1 - 2x)^(-1) + C.