If n is a positive integer, find the limit as n approaches +∞ of 1/n(sin (π/n) + sin (2π)/n + ... + sin (nπ)/n)
To find the limit as n approaches +∞ of the given expression, you can rewrite it using the Riemann sum formula for an integral.
Consider the function f(x) = sin(πx). We want to find the integral of f(x) from 0 to 1.
Using the Riemann sum formula, we have:
∫[0,1]f(x)dx ≈ (1/n)(f(0) + f(1/n) + f(2/n) + ... + f((n-1)/n))
In our case, we have:
1/n(sin(π/n) + sin(2π/n) + ... + sin(nπ/n))
This is similar to the Riemann sum formula, except that we have n values and an extra sin(0) term. We can rewrite it as:
(1/n)(sin(0) + sin(π/n) + sin(2π/n) + ... + sin(nπ/n))
Now, let's simplify this expression. The sin(0) term is 0, so it doesn't affect the limit as n approaches +∞. We are left with:
(1/n)(sin(π/n) + sin(2π/n) + ... + sin(nπ/n))
The next step is to recognize that this expression is now the Riemann sum for the integral of f(x) from 0 to 1.
Therefore, as n approaches +∞, this expression approaches the integral:
∫[0,1]f(x)dx
The integral of sin(πx) from 0 to 1 is [-cos(πx)] evaluated from 0 to 1:
[-cos(π(1))] - [-cos(π(0))]
Which simplifies to:
[-cos(π)] - [-cos(0)]
[-(-1)] - [-1]
1 - (-1)
1 + 1
2
Thus, the limit as n approaches +∞ of the original expression is 2.
To find the limit as n approaches positive infinity of the given expression, we can rewrite it using the concept of Riemann sums.
First, notice that each term in the expression can be written as sin(kπ/n), where k ranges from 1 to n. We can rewrite sin(kπ/n) as sin(π (k/n)).
Now, let's consider the sum ∑[k=1 to n] sin(π (k/n))/n. This is the sum of n identical terms, where each term is sin(π (k/n))/n.
To simplify further, we can recognize that the sum can be approximated by an integral. Consider the function f(x) = sin(πx)/x, where x ranges from 0 to 1.
The sum ∑[k=1 to n] sin(π (k/n))/n is roughly equal to the definite integral ∫[0 to 1] sin(πx)/x dx.
Now, let's evaluate the integral. Since it involves an improper integral, we need to take the limit as the upper bound approaches 1.
∫[0 to 1] sin(πx)/x dx = lim[δ→0] ∫[δ to 1] sin(πx)/x dx.
This integral is known as the Dirichlet integral and can be evaluated using techniques from calculus. The result of this integral is π/2.
Therefore, the limit as n approaches positive infinity of the given expression is π/2.