applying the integral test. use the integral test to determine if the series converge or diverge. be sure to…

Explanation:

Step1: Define the function

Let $f(x)=\frac{1}{x^{2}}$. This function is positive, continuous, and decreasing for $x\geq1$. The derivative $f^\prime(x)=-\frac{2}{x^{3}}<0$ for $x > 0$, so it is decreasing.

Step2: Set up the improper - integral

We consider the improper integral $\int_{1}^{\infty}\frac{1}{x^{2}}dx=\lim_{t\rightarrow\infty}\int_{1}^{t}x^{- 2}dx$.

Step3: Evaluate the integral

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\lim_{t\rightarrow\infty}\int_{1}^{t}x^{-2}dx=\lim_{t\rightarrow\infty}\left[- \frac{1}{x}\right]_{1}^{t}$.

Step4: Calculate the limit

$\lim_{t\rightarrow\infty}\left(-\frac{1}{t}+\frac{1}{1}\right)=0 + 1=1$. Since the improper integral $\int_{1}^{\infty}\frac{1}{x^{2}}dx$ converges, by the Integral Test, the series $\sum_{n = 1}^{\infty}\frac{1}{n^{2}}$ converges.

Answer:

The series $\sum_{n = 1}^{\infty}\frac{1}{n^{2}}$ converges.