Question

Use the estimate \(R_{N} \leq \int_{N}^{\infty} f(t) d t\) to find a bound for the remainder \(R_{N}=\sum_{n=1}^{\infty} a_{n}-\sum_{n=1}^{N} a_{n}\) where \(a_{n}=f(n)\). \(\sum_{n=1}^{1000} \frac{1}{n^{3}}\)

Short answer

The remainder is bounded by \( \frac{1}{2,000,000} \).

Step-by-step solution

Identify the Function

The first step is to identify the function represented by the series. Here, we are given that \( a_n = f(n) = \frac{1}{n^3} \). This function is the sequence term of the series \( \sum_{n=1}^{\infty} \frac{1}{n^3} \).

Rewrite the Remainder Bound Formula

Next, rewrite the remainder bound formula as \( R_N \leq \int_{N}^{\infty} f(t) \, dt \) using our function. For this problem, it becomes \( R_N \leq \int_{1000}^{\infty} \frac{1}{t^3} \, dt \) since we're summing up to \( N = 1000 \).

Set Up the Integral

Set up the improper integral \( \int_{1000}^{\infty} \frac{1}{t^3} \, dt \). This integral will be solved to find the bound on \( R_N \).

Solve the Integral

The antiderivative of \( \frac{1}{t^3} \) is \( -\frac{1}{2t^2} \). Evaluate the improper integral from 1000 to \( \infty \).\[ \int_{1000}^{\infty} \frac{1}{t^3} \, dt = \lim_{b \to \infty} \left[ -\frac{1}{2t^2} \right]_{1000}^{b} \] \[ = \lim_{b \to \infty} \left(-\frac{1}{2b^2} + \frac{1}{2(1000)^2}\right) \]As \( b \to \infty \), \(-\frac{1}{2b^2} \to 0\). So,\[ R_{1000} \leq \frac{1}{2 \times 1000^2} = \frac{1}{2,000,000} \].

Conclusion

We have computed the integral to find the remainder bound: \( R_{1000} \leq \frac{1}{2,000,000} \). Thus, the bound for the remainder of the series \( \sum_{n=1}^{\infty} \frac{1}{n^3} \) after the 1000th term is \( \frac{1}{2,000,000} \).

Key concepts

Improper Integrals

Improper integrals extend the concept of definite integrals to include intervals that are infinitely long or have unbounded functions. They are essential when dealing with functions that approach infinity. In these scenarios, we can't utilize standard integration methods since either the limits of integration or the function itself could be infinite. Instead, we use limits to express the idea of the integral.
For instance, when calculating the improper integral \( \int_{1000}^{\infty} \frac{1}{t^3} \, dt \), we replace the upper bound of infinity with a variable, like \( b \), and take the limit as \( b \to \infty \). Then, the integral is solved from 1000 to \( b \), allowing us to evaluate a definite value even when dealing with something seemingly boundless.
Such improper integrals are crucial when estimating series remainders because they help provide bounds for the sums that otherwise might be very challenging to calculate directly. It's a way of understanding the behavior of the series far beyond the finite cut-off point.

Convergence of Series

The convergence of an infinite series is a central topic in mathematical analysis. It determines whether a series approaches a finite limit as more terms are added, or whether it tends towards infinity or oscillates indefinitely.
The convergence of a series like \( \sum_{n=1}^{\infty} \frac{1}{n^3} \) can be tested using various methods, such as the integral test. This specific series is known as a p-series with \( p = 3 \). The integral test states that if \( f(n) = a_n \) is positive, continuous, and decreasing, then the series \( \sum_{n=1}^{\infty} a_n \) converges if the improper integral \( \int_{1}^{\infty} f(x) \, dx \) also converges.
If the integral converges, it implies that the original series also converges, thus confirming the sum of its infinite terms approaches a finite number. Strong understanding of series convergence is crucial for working out practical problems in calculus and analysis.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are the reverse process of differentiation. Finding the antiderivative involves determining which function, when differentiated, leads to the integrand. In practical terms, having an antiderivative is vital for evaluating integrals and by extension improper integrals, which help bound series remainders.
For the function \( \frac{1}{t^3} \), we find its antiderivative by reversing the process of differentiation, yielding \( -\frac{1}{2t^2} \). This is key to solving \( \int_{1000}^{\infty} \frac{1}{t^3} \, dt \), allowing us to evaluate the improper integral by using limits and obtain a finite value.
Knowing how to compute antiderivatives allows us to understand the accumulation of quantities and compute areas and limits, making it a powerful tool in calculus.