급수 심화 전체 문제

(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 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.

(a) Differentiating both sides of

$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

gives

$$\sum_{n=0}^\infty n x^{n-1}=\frac{1}{(1-x)^2}$$

Now multiplying both sides by $x$ gives

$$\sum_{n=0}^\infty n x^n=\frac{x}{(1-x)^2}$$

as desired.

(b) Differentiating both sides of the conclusion of part (a) gives

$$\sum_{n=0}^\infty n^2 x^{n-1}=\frac{(1-x)^2-2x(x-1)}{(1-x)^4} =\frac{(1-x)(1-x+2x)}{(1-x)^4} =\frac{1+x}{(1-x)^3}$$

Now multiplying both sides by $x$ gives

$$\sum_{n=0}^\infty n^2 x^n=\frac{x(1+x)}{(1-x)^3}$$

We know that differentiation preserves the radius of convergence of power series. So this series has radius of convergence $1$ (the radius of convergence of the original geometric series). At $x=\pm1$ the series diverges by the divergence test. So the series converges for $-1<x<1$.

(a) We know that the radius of convergence $R$ obeys

$$\begin{aligned} \frac{1}{R} = \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty}\frac{n}{n+1}\ \frac{(n+1)a_{n+1}}{na_n} = 1 \frac{C}{C} = 1 \end{aligned}$$

because we are told that $\lim\limits_{n\rightarrow\infty} na_n=C$. So $R=1$.

(b) Just knowing that the radius of convergence is $1$, we know that the series converges for $|x|<1$ and diverges for $|x|>1$. That leaves $x\pm 1$.

When $x=+1$, the series reduces to $\sum\limits_{n=1}^\infty a_n$. We are told that $na_n$ decreases to $C>0$. So $a_n\ge \frac{C}{n}$. By the comparison test with the harmonic series $\sum\limits_{n=1}^\infty\frac{1}{n}$, which diverges by the $p$–test with $p=1$, our series diverges when $x=1$.

When $x=-1$, the series reduces to $\sum\limits_{n=1}^\infty (-1)^n a_n$. We are told that $na_n$ decreases to $C>0$. So $a_n>0$ and $a_n$ converges to $0$ as $n\rightarrow\infty$. Consequently $\sum\limits_{n=1}^\infty (-1)^n a_n$ converges by the alternating series test.

In conclusion $\sum\limits_{n=1}^\infty a_n x^n$ converges when $-1\le x< 1$.

First, we differentiate.

$$\begin{aligned} f(x)&=\displaystyle\sum_{n=0}^{\infty}A_n(x-c)^n\\ f'(x)&=\sum_{n=0}^{\infty}nA_n(x-c)^{n-1}\\ &=\sum_{n=1}^{\infty}nA_n(x-c)^{n-1}\\ f'(c)&=\sum_{n=1}^\infty nA_n\cdot 0^{n-1}\\ &=A_1\cdot 1 + 2A_2\cdot0+3A_3\cdot0+\cdots\\ &=A_1 \end{aligned}$$

So, if $A_1=0$, then $f'(c)=0$. That is, $f(x)$ has a critical point at $x=c$.

To determine the behaviour of this critical point, we use the second derivative test.

$$\begin{aligned} f'(x)&=\sum_{n=1}^{\infty}nA_n(x-c)^{n-1}\\ f''(x)&=\sum_{n=1}^{\infty}n(n-1)A_n(x-c)^{n-2}\\ &=\sum_{n=2}^{\infty}n(n-1)A_n(x-c)^{n-2}\\ f''(c)&=\sum_{n=2}^{\infty}n(n-1)A_n\cdot 0^{n-2}\\ &=2(1)A_2\cdot 0^0+3(2)A_3\cdot 0^1+4(3)A_4\cdot 0^2+\cdots\\ &=2A_2 \end{aligned}$$

Following the second derivative test, $x=c$ is the location of a local maximum if $A_2<0$, and it is the location of a local minimum if $A_2>0$. (If $A_2=0$, the critical point may or may not be a local extremum.)

The factor $(n+1)(n+2)$ reminds us of a derivative. Start with the geometric series.

$$\begin{aligned} \frac{1}{1-x}&=\sum_{n=0}^\infty x^n\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{1-x} \right\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \sum_{n=0}^\infty x^n \right\}\\ \frac{1}{(1-x)^2}&=\sum_{n=0}^\infty nx^{n-1}=\sum_{n=1}^\infty nx^{n-1}\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{(1-x)^2} \right\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{ \sum_{n=1}^\infty nx^{n-1} \right\}\\ \frac{2}{(1-x)^3}&=\sum_{n=1}^\infty n(n-1)x^{n-2}=\sum_{n=2}^\infty n(n-1)x^{n-2}\\ &=\sum_{n=0}^\infty (n+2)(n+1)x^n \end{aligned}$$

Let $x=\frac{1}{7}$. Then $|x|<1$, so our series converges.

$$\begin{aligned} \frac{2}{(1-1/7)^3} &=\sum_{n=0}^\infty (n+2)(n+1)\left(\frac{1}{7}\right)^n\\ \frac{2}{(6/7)^3} &=\sum_{n=0}^\infty \frac{(n+2)(n+1)}{7^n} \end{aligned}$$