급수 심화 전체 문제

This is a geometric series with $r=1.001$. Since $|r|>1$, it is divergent.

This is a geometric series with $r=\frac{-1}{5}$. Since $|r|<1$, it is convergent.

We want to use the formula $\sum_{n=0}^\infty r^n=\frac{1}{1-r}$, but our series does not start at 0, so we re-write it:

$$\begin{aligned} \sum_{n=3}^\infty \left(\frac{-1}{5}\right)^n&= \sum_{n=0}^\infty \left(\frac{-1}{5}\right)^n- \sum_{n=0}^2 \left(\frac{-1}{5}\right)^n =\frac{1}{1-(-1/5)} - \left(1-\frac{1}{5}+\frac{1}{25} \right)\\ &=\frac{1}{6/5}-1+\frac{1}{5}-\frac{1}{25} = \frac{5}{6}+\frac{-25+5-1}{25}=-\frac{1}{150} \end{aligned}$$

This is close to being in the form of a geometric series. First, we should have our powers be $k$, not $k+2$, but we notice $3^{k+2}=3^k3^2=9\cdot3^2$, so:

$$\begin{aligned} \sum_{k=0}^\infty\frac{2^k}{3^{k+2}}&=\sum_{k=0}^\infty\frac{2^k}{9\cdot3^k} =\frac{1}{9}\sum_{k=0}^\infty\frac{2^k}{3^k}=\frac{1}{9}\sum_{k=0}^\infty\left(\frac{2}{3}\right)^k \\ \text{Now it looks like a geometric series with $r=\frac{2}{3}$} \\ &=\frac{1}{9}\left(\frac{1}{1-(2/3)}\right)=\frac{1}{3} \end{aligned}$$

In conclusion: this (geometric) series is convergent, and its sum is $\dfrac{1}{3}$.

$$\begin{aligned} \sum_{n=5}^\infty \frac{1}{e^n}&=\sum_{n=5}^\infty \left(\frac{1}{e}\right)^n\\ &=\sum_{n=0}^\infty \left(\frac{1}{e}\right)^n-\sum_{n=0}^4 \left(\frac{1}{e}\right)^n\\ &=\dfrac{1}{1-\frac{1}{e}}-\dfrac{1-\left(\frac{1}{e}\right)^{5}}{1-\frac{1}{e}}\\ &=\dfrac{\left(\frac{1}{e}\right)^5}{1-\frac{1}{e}} =\dfrac{1}{e^5\left(1-\frac{1}{e}\right)}\\ &=\dfrac{1}{e^5-e^4} \end{aligned}$$

The series $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{9n+5}$ converges by the alternating series test. On the other hand the series $\sum\limits_{n=1}^{\infty}\big|\frac{(-1)^n}{9n+5}\big| =\sum_{n=1}^{\infty}\frac{1}{9n+5}$ diverges by the limit comparison test with $b_n=\frac{1}{n}$. So the given series is conditionally convergent.

The sequence has some positive terms and some negative terms, which limits the tests we can use. However, if we consider the series $\displaystyle\sum_{n=1}^\infty \left|\dfrac{\sin n}{n^2}\right|$, we can use the direct comparison test.

For every $n$, $| \sin n| <1$, so $0\le \left|\dfrac{\sin n}{n^2}\right|<\dfrac{1}{n^2}$. Since $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}$ converges, then by the direct comparison test, $\displaystyle\sum_{n=1}^\infty \left|\dfrac{\sin n}{n^2}\right|$ converges as well. Then $\displaystyle\sum_{n=1}^\infty \dfrac{\sin n}{n^2}$ converges absolutely– in particular, it converges.

The terms of this series are sometimes negative and sometimes positive. But, they are not strictly alternating, so we can't use the alternating series test. Instead, we use a direct comparison test to show the series converges absolutely.

$$\begin{aligned} \left|\frac{\sin^2 n - \cos^2 n+\tfrac12}{2^n}\right| \le\frac{1 + 1 +\tfrac12}{2^n} =\frac{5}{2^{n+1}} \end{aligned}$$

The series $\displaystyle\sum_{n=1}^\infty \frac{5}{2^{n+1}}$ converges, because it's a geometric series with $r=\frac{1}{2}$. By the direct comparison test, $\displaystyle\sum_{n=1}^\infty \left|\frac{\sin^2 n - \cos^2 n+\tfrac12}{2^n}\right|$ converges as well. Then $\displaystyle\sum_{n=1}^\infty \frac{\sin^2 n - \cos^2 n+\tfrac12}{2^n}$ converges absolutely, so it converges.