To bound the remainder $R_4 = \sum_{n=5}^{\infty} \frac{1}{n^2}$ using the integral test, we need to analyze the convergence and bounds of the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$, specifically from $n = 5$ onward.

The integral test tells us that if $f(x) = \frac{1}{x^2}$ is positive, continuous, and decreasing for $x \geq 5$, then the series $\sum_{n=5}^{\infty} \frac{1}{n^2}$ can be bounded by the integral:

$\int_{5}^{\infty} \frac{1}{x^2} \, dx$

Steps:

1. Compute the Integral for the Upper Bound: $\int_{5}^{\infty} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_{5}^{\infty} = 0 - \left(-\frac{1}{5}\right) = \frac{1}{5}$

   This result tells us that the series remainder $R_4$ is bounded above by $\frac{1}{5}$.

2. Compute a Lower Bound: Since the series is decreasing, a lower bound for the sum $R_4$ can be approximated by the next integral: $\int_{6}^{\infty} \frac{1}{x^2} \, dx = \frac{1}{6}$

3. Conclusion: Therefore, we have the following bounds for $R_4$: $\frac{1}{6} < R_4 < \frac{1}{5}$

These bounds provide an estimate for the remainder of the series starting from $n = 5$ onward.

Would you like further details or have any questions on this solution?

Related Questions:

1. How does the integral test help in determining the convergence of a series?
2. What are other methods to bound the remainder of a convergent series?
3. How do we interpret the results from the integral test in terms of convergence speed?
4. Could we use a comparison test to find similar bounds for this series?
5. How does the choice of the lower limit (starting index) affect the remainder estimates?

Tip:

When using the integral test for series convergence, remember that the integral provides a continuous approximation of the sum, which is particularly useful for bounding remainders.