Explain clearly in detail the difference of the summation of n=1 to infinity of 1/n^2; and the integral of 1/x^2dx with the limit of 1 to infinity, as well.
Part 2: Explain the natures of the two expressions and explain why the summation is greater than the integral test.
The sum=1.645
Integral=1
Any help would be greatly appreciated
Drawa graph of the function
f(x) = 1/x^2.
The summation
f(1) + f(2) + f(3) + ...
can be intepreted as the total area of the rectangles of heights f(1), f(2),
f(3),... and all of width 1.
If you draw these rectangles in your graph, the one with height f(1) is between x = 1 and x = 2, the rectangle of height f(2) is between x = 2 and x = 3, etc., then you clearly see that the total area of all the rectangles together is less than the area under the curve, due to the fact that f(x) is a decreasing function.
THe precise relation between the summation and the integration is given by the Euler–Maclaurin summation formula:
http://en.wikipedia.org/wiki/Euler–Maclaurin_formula
To understand the difference between the summation and the integral, let's first examine the properties of these two mathematical concepts.
1. Summation:
The summation, denoted by the Greek capital letter sigma (Σ), represents the addition of a sequence of terms. In this case, the summation of n=1 to infinity of 1/n^2 is given by:
S = 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ...
This expression represents the sum of infinitely many terms, where each term is the reciprocal of the square of n.
2. Integral:
The integral, denoted by the long "S" symbol (∫), is a mathematical tool used to calculate the area under a curve. In this case, the integral of 1/x^2dx with the limit of 1 to infinity represents the integral of the function f(x) = 1/x^2 over an infinite range:
I = ∫(1/x^2)dx, with the limit of 1 to infinity
This expression calculates the area under the curve of the function f(x) = 1/x^2 from x = 1 to x = infinity.
Now, let's discuss the nature of these two expressions and why the summation is greater than the integral.
1. Summation:
For the given summation Σ(1/n^2), the terms in the series are positive and decreasing. As n increases, the individual terms become smaller since the denominator increases faster than the numerator. This guarantees convergence, which means that the sum of the terms approaches a certain finite value. In this case, the sum of the series is approximately 1.645.
2. Integral:
The integral of the function f(x) = 1/x^2 is equal to -1/x evaluated from the lower limit of 1 to the upper limit of infinity. Calculating the integral, we get:
∫(1/x^2)dx = [-1/x] evaluated from 1 to infinity
Substituting the limits, we have:
I = [-1/infinity] - [-1/1] = 0 - (-1) = 1
The integral of 1/x^2 with the limit of 1 to infinity equals 1.
The reason the summation is greater than the integral in this case is due to the different mathematical properties of these two concepts. The integral calculates the area under a curve, and even though the function 1/x^2 decreases as x increases, the area under the curve is finite and equals 1. On the other hand, the summation adds infinitely many decreasing terms, which converge to a finite value of approximately 1.645.
In summary, the summation Σ(1/n^2) converges to approximately 1.645, while the integral of 1/x^2 with the limit of 1 to infinity equals 1.