Convergence Divergence

Let's use the Limit Comparison test . It says: let $\displaystyle \sum_{n=1}^{\infty} a_n$ and $\displaystyle \sum_{n=1}^{\infty} b_n$ be series of positive terms. Then:

• If $\displaystyle \lim_{n \to \infty} \dfrac{a_n}{b_n} > 0$ and it's finite, $\displaystyle \sum_{n=1}^{\infty} a_n$ converges if and only if $\displaystyle \sum_{n=1}^{\infty} b_n$ converges.
• If $\displaystyle \lim_{n \to \infty} \dfrac{a_n}{b_n} = 0$ and $\displaystyle \sum_{n=1}^{\infty} b_n$ converges, then $\displaystyle \sum_{n=1}^{\infty} a_n$ converges.
• If $\displaystyle \lim_{n \to \infty} \dfrac{a_n}{b_n} = \infty$ and $\displaystyle \sum_{n=1}^{\infty} b_n$ diverges, then $\displaystyle \sum_{n=1}^{\infty} a_n$ diverges.

Now, let's determine the convergence of each series:

A . Choose the series $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n}$ because we know that diverges. We have $\displaystyle \lim_{n \to \infty} \dfrac{\sin \frac{1}{n}}{\frac{1}{n}}=\lim_{n \to 0} \dfrac{\sin n}{n}=1>0$ , so this series diverges .

B . Choose the series $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2}$ because we know that converges. We have $\displaystyle \lim_{n \to \infty} \dfrac{\sin \frac{1}{n^2}}{\frac{1}{n^2}}=\lim_{n \to \infty} \dfrac{-\frac{2}{n^3} \cos \frac{1}{n^2}}{-\frac{2}{n^3}}=\lim_{n \to \infty} \cos \dfrac{1}{n^2}=1>0$ , so this series converges .

C . Again choose the series $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n}$ . We have $\displaystyle \lim_{n \to \infty} \dfrac{\cos \frac{1}{n}}{\frac{1}{n}}=\lim_{n \to 0} \dfrac{\cos n}{n}=\infty$ , so this series diverges .

D . Again choose $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n}$ . We have $\displaystyle \lim_{n \to \infty} \dfrac{\cos \frac{1}{n^2}}{\frac{1}{n}}=\lim_{n \to \infty} n \cos \dfrac{1}{n^2}=\infty$ , so this series diverges .

So, the false statement is $\boxed{D}$ .

Since for very small x cos(x) is approximately 1, you can instantly see that D has to be wrong, because for a big enough n the term $\frac{1}{n^2}$ is close enough to 0 that you are pretty much adding up an infinite number of 1's, which definitely doesn't converge