급수 심화 전체 문제

Note that $(-1)^{2n+1}=(-1)\cdot(-1)^{2n}=-1$. So we can simplify

$$\begin{aligned} \sum_{n=1}^\infty\frac{(-1)^{2n+1}}{1+n}=-\sum_{n=1}^\infty\frac{1}{1+n} \end{aligned}$$

Since $\displaystyle \frac{1}{1+n} \geq \frac{1}{n+n} = \frac{1}{2n}$, $\ \ \displaystyle \sum_{n=1}^\infty\frac{1}{1+n}$ diverges by the comparison test with the divergent harmonic series $\sum\limits_{n=1}^\infty\frac{1}{n}$. The extra overall factor of $-1$ in the original series does not change the conclusion of divergence.

Since

$$\begin{aligned} \lim_{n\to\infty} \frac{1+4^n}{3+2^{2n}} =\lim_{n\to\infty} \frac{1+4^n}{3+4^n} = 1 \end{aligned}$$

the alternating series test cannot be used. Indeed, $\displaystyle \lim_{n\to\infty} (-1)^{n-1} \frac{1+4^n}{3+2^{2n}}$ does not exist (for very large $n$, $(-1)^{n-1} \frac{1+4^n}{3+2^{2n}}$ alternates between a number close to $+1$ and a number close to $-1$) so the divergence test says that the series diverges. (Note that “none of the above” cannot possibly be the correct answer — every series either converges absolutely, converges conditionally, or diverges.)

We first develop some intuition about $\displaystyle\sum_{n=1}^\infty\left|\frac{n^2-\sin n}{n^6+n^2}\right|$, where we take the absolute value of the summands to consider whether the series converges absolutely. For very large $n$, $n^2$ dominates $\sin n$ and $n^6$ dominates $n^2$ so that

$$\begin{aligned} \left|\frac{n^2-\sin n}{n^6+n^2}\right| \approx \frac{n^2}{n^6} =\frac{1}{n^4} \end{aligned}$$

The series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^4}$ converges by the $p$–test with $p=4>1$. We expect the given series to converge too.

To verify that our intuition is correct, we apply the limit comparison test with

$$\begin{aligned} a_n= \frac{n^2-\sin n}{n^6+n^2} \quad\text{and}\quad b_n= \frac{1}{n^4} \end{aligned}$$

which is valid since

$$\lim_{n\rightarrow\infty} \frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\left|\frac{(n^2-\sin n)}{n^6+n^2}\right|\cdot\frac{n^4}{1} =\lim_{n\rightarrow\infty}\frac{|n^6-n^4\sin n|}{n^6+n^2} =\lim_{n\rightarrow\infty}\frac{1-n^{-2}\sin n}{1+n^{-4}} =1$$

exists and is nonzero. Since the series $\sum\limits_{n=1}^\infty b_n$ converges, the series $\sum\limits_{n=1}^\infty\dfrac{|n^2-\sin n|}{n^6+n^2}$ converges too. Therefore, the series $\sum\limits_{n=1}^\infty\dfrac{n^2-\sin n}{n^6+n^2}$ converges absolutely.

You might think that this series converges by the alternating series test. But you would be wrong. The problem is that $\{a_n\}$ does not converge to zero as $n\rightarrow\infty$, so that the series actually diverges by the divergence test. To verify that the $n^{\text{th}}$ term does not converge to zero as $n\rightarrow\infty$ let's write $a_n= \frac{(2n)!}{(n^2+1)(n!)^2}$ (i.e. $a_n$ is the $n^{\text{th}}$ term without the sign) and check to see whether $a_{n+1}$ is bigger than or smaller than $a_n$.

$$\begin{aligned} \frac{a_{n+1}}{a_n} &=\frac{(2n+2)!}{((n+1)^2+1)((n+1)!)^2} \frac{(n^2+1)(n!)^2}{(2n)!} =\frac{(2n+2)(2n+1)}{(n+1)^2}\frac{n^2+1}{(n+1)^2+1} \\ &=\frac{2(2n+1)}{(n+1)}\frac{1+1/n^2}{(1+1/n)^2+1/n^2} =4\frac{1+1/2n}{1+1/n}\frac{1+1/n^2}{(1+1/n)^2+1/n^2} \end{aligned}$$

So

$$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=4$$

and, in particular, for large $n$, $a_{n+1}>a_n$. Thus, for large $n$, $a_n$ increases with $n$ and so cannot converge to $0$. So the series diverges by the divergence test.

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.

(a)

• Solution 1: We need to show that $\sum\limits_{n=1}^\infty 24n^2 e^{-n^3}$ converges. If we replace $n$ by $x$ in the summand, we get $f(x) = 24x^2 e^{-x^3}$, which we can integate. (Just substitute $u=x^3$.) So let's try the integral test. First, we have to check that $f(x)$ is positive and decreasing. It is certainly positive. To determine if it is decreasing, we compute

$$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}x} = 48x e^{-x^3} - 24\times 3 x^4 e^{-x^3} = 24x (2-3x^3) e^{-x^3} \end{aligned}$$

which is negative for $x\ge1$. Therefore $f(x)$ is decreasing for $x\ge1$, and the integral test applies. The substitution $u=x^3$, $\mathrm{d}u=3x^2\,\,\mathrm{d}x$, yields

$$\begin{aligned} \int f(x) \,\mathrm{d}x = \int 24x^2 e^{-x^3} \,\mathrm{d}x = \int 8 e^{-u}\,\mathrm{d}u = -8e^{-u} + C = -8e^{-x^3} + C. \end{aligned}$$

Therefore

$$\begin{aligned} \int_1^\infty f(x) \,\mathrm{d}x &= \lim_{R \to \infty} \int_1^R f(x) \,\mathrm{d}x = \lim_{R \to \infty} \bigg[ {-}8 e^{-x^3} \bigg]_1^R \\ &= \lim_{R \to \infty} ( - 8 e^{-R^3} + 8 e^{-1} ) = 8 e^{-1} \end{aligned}$$

Since the integral is convergent, the series $\sum\limits_{n=1}^\infty 24n^2 e^{-n^3}$ converges and the series $\displaystyle \sum_{n=1}^\infty (-1)^{n-1}24n^2 e^{-n^3}$ converges absolutely.

• Solution 2: Alternatively, we can use the ratio test with $a_n=24n^2 e^{-n^3}$. We calculate

$$\begin{aligned} \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty} \left|\frac{24 (n+1)^2e^{-(n+1)^3}}{24 n^2e^{-n^3}} \right| \\ &= \lim_{n\to\infty} \left( \frac{(n+1)^2}{n^2} \frac{e^{n^3}}{e^{(n+1)^3}} \right) \\ &= \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^2 e^{-(3n^2+3n+1)} = 1\cdot0=0 < 1, \end{aligned}$$

and therefore the series converges absolutely.

• Solution 3: Alternatively, alternatively, we can use the limiting comparison test. First a little intuition building. Recall that we need to show that $\sum\limits_{n=1}^\infty 24n^2 e^{-n^3}$ converges. The $n^{\rm th}$ term in this series is

$$a_n = 24n^2 e^{-n^3} =\frac{24 n^2}{e^{n^3}}$$

It is a ratio with both the numerator and denominator growing with $n$. A good rule of thumb is that exponentials grow a lot faster than powers. For example, if $n=10$ the numerator is $2400=2.4\times 10^3$ and the denominator is about $2\times 10^{434}$. So we would guess that $a_n$ tends to zero as $n\rightarrow\infty$. The question is “does $a_n$ tend to zero fast enough with $n$ that our series converges?”. For example, we know that $\sum_{n=1}^\infty \frac{1}{n^2}$ converges (by the $p$–test with $p=2$). So if $a_n$ tends to zero faster than $\frac{1}{n^2}$ does, our series will converge. So let's try the limiting convergence test with $a_n = 24n^2 e^{-n^3} =\frac{24 n^2}{e^{n^3}}$ and $b_n=\frac{1}{n^2}$.

$$\begin{aligned} \lim_{n\rightarrow\infty}\frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{24n^2 e^{-n^3}}{1/n^2} =\lim_{n\rightarrow\infty}\frac{24n^4 }{e^{n^3}} \end{aligned}$$

By l'Hopital's rule, twice,

$$\begin{aligned} \lim_{x\rightarrow\infty} \frac{24x^4 }{e^{x^3}} &=\lim_{x\rightarrow\infty} \frac{4\times 24x^3 }{3 x^2e^{x^3}} &\text{by l'H\^opital} \\ &=\lim_{x\rightarrow\infty} \frac{32x }{e^{x^3}} &\text{just cleaning up} \\ &=\lim_{x\rightarrow\infty} \frac{32 }{3x^2e^{x^3}} &\text{by l'H\^opital, again} \\ &=0 \end{aligned}$$

That's it. The limit comparison test now tells us that $\sum_{n=1}^\infty a_n$ converges.

(b) In part (a) we saw that $24n^2 e^{-n^3}$ is positive and decreasing. The limit of this sequence equals $0$ (as can be shown with l'Hopital's Rule, just as we did at the end of the third solution of part (a)). Therefore, we can use the alternating series test, so that the error made in approximating the infinite sum $S= \sum\limits_{n=1}^\infty a_n = \sum\limits_{n=1}^\infty (-1)^{n-1} 24n^2 e^{-n^3}$ by the sum of its first $N$ terms, $S_N=\sum\limits_{n=1}^N a_n$, lies between $0$ and the first omitted term, $a_{N+1}$. If we use $5$ terms, the error satisfies

$$\begin{aligned} |S - S_5 | \le |a_6| = 24 \times 36 e^{-6^3}\approx 1.3 \times 10^{-91} \end{aligned}$$

The error in our approximation using through term $N$ is at most $\frac{1}{(2(N+1))!}$. We want $\frac{1}{(2(N+1))!}<\frac{1}{1000}$. By checking small values of $N$, we see that $8!=40320>1000$, so if $N=3$, then $\frac{1}{2(N+1)!}=\frac{1}{40320}<\frac{1}{1000}$. So, for our approximation, it suffices to consider the first four terms of our series.

$$\begin{aligned} \cos(1)&\approx \sum_{N=0}^3 \frac{(-1)^n}{(2n)!} = \frac{1}{0!}-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}\\ &=1-\frac12+\frac{1}{24}-\frac{1}{720}\\ &=\frac{720-360+30-1}{720}=\frac{389}{720} \end{aligned}$$

When we use a calculator, we see

$$\begin{aligned} \frac{389}{720}&=0.5402\overline{77} \\ \cos(1)&\approx 0.540302\\ \cos(1) - \frac{389}{720}&\approx 0.000024528\approx \frac{1}{40770} \end{aligned}$$

So, our error is reasonably close to our bound of $\frac{1}{40320}$, and far smaller than $\frac{1}{1000}$.

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.

If $n$ is prime, then $$\left| \frac{a_n}{e^n}\right|=\left|-\frac{e^{n/2}}{e^n}\right|=\frac{1}{e^{n/2}}=\left(\frac{1}{\sqrt e}\right)^n$$

If $n$ is not prime, then $$\left| \frac{a_n}{e^n}\right|=\left|-\frac{n^2}{e^n}\right|=\frac{n^2}{e^n}$$ For $n$ sufficiently large, $n^2<e^{n/2}$, so for $n$ sufficiently large, $$\frac{n^2}{e^n}\le \left(\frac{1}{\sqrt{e}}\right)^n.$$

Since $e>1$, then $\sqrt{e}>1$, so the geometric series $\displaystyle\sum \left(\frac{1}{\sqrt{e}}\right)^n$ has $|r|=r=\frac{1}{\sqrt{e}}<1$, so it converges. By the direct comparison test, $\displaystyle\sum_{n=1}^\infty\left|\dfrac{a_n}{e^n}\right|$ converges as well. Then $\displaystyle\sum_{n=1}^\infty\dfrac{a_n}{e^n}$ converges absolutely, so it converges.