An archived instance of discourse for discussion in undergraduate Real and Numerical Analysis.

# Quiz problem: Estimate with the integral test

mark

Suppose we want to estimate
$$\sum_{k=1}^{\infty} \frac{n}{n^5+1}$$
with a finite sum and we'd like the answer to be within $0.0001$ of the actual value.
How many terms do we need?

XCNate

It seems like the simplest thing to do would be to decide the number of terms needed to estimate $$\sum _{k=1}^\infty \frac{1}{n^4}$$ since $\frac{n}{n^5 + 1} < \frac{1}{n^4}$ for $n > 0.$

mark

@XCNate Yes, good observation.

Professor.Membrane

building off @XCNate's post we have the following:
$$\sum_{n=N}^{\infty} \frac{n}{n^{5} + 1} \leq \int_{N}^{\infty} \frac{1}{x^4} dx \leq \epsilon = .0001.$$
Note that:
$$\int_{N}^{\infty} \frac{1}{x^4} dx = \frac{-1}{3x^3} \Bigg|_{N}^{\infty} = 0 - \frac{-1}{3N^{3}} = \frac{1}{3N^{3}} \leq \epsilon = .0001$$
Hence we choose $N$ to satisfy:
$$N \geq \sqrt[3]{\frac{1}{3(.0001)}} \approx 15$$

mark

@Professor.Membrane Very close! Keep in mind that theorem 2.2 of our text has states that
$$\int_N^{\infty} f(x) dx \leq \sum_{n=N}^{\infty} a_n \leq f(N) + \int_N^{\infty} f(x)dx.$$
So, you've got that little extra $f(N)$ in the upper bound. I think that 16 should actually work.