급수 심화 전체 문제

Set $f(x)=(2+x)^3$, so we are approximating $f(0.001)$.

The obvious choice of $a$ is $a=0$.

Then $f'(x)=3(2+x)^2$, so $$(2.001)^3=f(0.001)\approx f(0)+f'(0)(0.001-0) =8+\frac{12}{1000}=\frac{8012}{1000}$$

Remark: if we had chosen $f(x)=x^3$ and $a=2$, the result would have been exactly the same.

We set $f(x)=(8+x)^{2/3}$, and choose $a=0$ as our centre.

Then $f'(x)=\dfrac{2}{3}(8+x)^{-1/3}$, so that

$$\begin{aligned} (8.06)^{2/3}=f(0.06)&\approx f(0)+f'(0)\cdot 0.06\\ &=8^{2/3}+\frac{2}{3}8^{-1/3}\cdot 0.06\\ &=\sqrt[3]{8}^2+\frac{2}{3\sqrt[3]{8}}\cdot 0.06\\ &=2^2+\frac{2}{3\cdot 2}\cdot 0.06\\ &=4+\frac{1}{3}\cdot 0.06\\&=4.02=\frac{402}{100}\end{aligned}$$

We begin by finding the derivatives of $f$ at $x=0$.

$$\begin{aligned} f(x)&=(1 - 3x)^{-1/3} & f(0) & = 1 \\ f'(x)&=(-3)\frac{-1}{3}(1 - 3x)^{-4/3}=(1-3x)^{-4/3} & f'(0) & = 1 \\ f''(x)&=(-3)\frac{-4}{3}(1-3x)^{-7/3}={4}(1 - 3x)^{-7/3} & f''(0) & = 4 \\ f^{(3)}(x)&=(-3)(4)\frac{-7}{3}(1-3x)^{-10/3}=28(1 - 3x)^{-10/3} & f^{(3)}(0) & = 28 \\ \end{aligned}$$

Plugging these into the definition of a Taylor Polynomial, we find that the third–order Taylor polynomial for $f$ around $x = 0$ is

$$\begin{aligned} T_3(x)&=1+x+\frac{4}{2!}x^2+\frac{28}{3!}x^3\\ &=1 + x +2x^2+\frac{14}{3}x^3 \end{aligned}$$

First, we'll check the divergence test. It doesn't always work, but if it does, it's likely the easiest path.

$$\begin{aligned} \lim_{n\to\infty} \frac{n^2}{3n^2+\sqrt n} \left(\frac{\frac{1}{n^2}}{\frac{1}{n^2}}\right) = \lim_{n\to\infty} \frac{1}{3+\frac{1}{n\sqrt n}} = \frac{1}{3} \neq 0 \end{aligned}$$

Since the limit of the terms being added is not zero, the series diverges by the divergence test.

This precise question was asked on a 2014 final exam. Note that the $n^{\rm th}$ term in the series is $a_n=\frac{5^k}{4^k+3^k}$ and does not depend on $n$! There are two possibilities. Either this was intentional (and the instructor was being particularly nasty) or it was a typo and the intention was to have $a_n=\frac{5^n}{4^n+3^n}$. In both cases, the limit

$$\begin{aligned} \lim_{n\to\infty} a_n &=\lim_{n\to\infty} \frac{5^k}{4^k+3^k} = \frac{5^k}{4^k+3^k} \neq 0 \\ \lim_{n\to\infty} a_n &=\lim_{n\to\infty} \frac{5^n}{4^n+3^n} =\lim_{n\to\infty} \frac{(5/4)^n}{1+(3/4)^n} = +\infty \neq 0 \\ \end{aligned}$$

is nonzero, so the series diverges by the divergence test.

The terms of the series tend to 0, so we can't use the divergence test.

To generate a guess about its convergence, we do the following: $$\sum \frac{1}{\sqrt{k}\sqrt{k+1}}=\sum\frac{1}{\sqrt{k^2+k}} \approx \sum\frac{1}{\sqrt{k^2}}=\sum\frac{1}{k}$$We guess that our series behaves like the harmonic series, and the harmonic series diverges (which can be demonstrated by $p$-test or integral test). So, we guess that our series diverges. However, in order to directly compare our series to the harmonic series and show our series diverges, our terms would have to be bigger than the terms in the harmonic series, and this is not the case. So, we use limit comparison.

$$\begin{aligned} \frac{\frac{1}{k}}{\frac{1}{\sqrt{k^2+k}}}&=\frac{\sqrt{k^2+k}}{k}=\frac{\sqrt{k^2+k}}{\sqrt{k^2}}= \sqrt{\frac{k^2+k}{k^2}}=\sqrt{1+\frac{1}{k}},\text{ so}\\ \lim_{k \rightarrow \infty}\frac{\frac{1}{k}}{\frac{1}{\sqrt{k^2+k}}}&=\lim_{k \rightarrow \infty}\sqrt{1+\frac{1}{k}}=1 \end{aligned}$$

Since 1 is a real number greater than 0, by the Limit Comparison Test, $\sum \frac{1}{\sqrt{k}\sqrt{k+1}}$ diverges, like $\sum \frac{1}{k}$.

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:

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

For any integer $n$, $\sin(\pi n)=0$, so $\sum \sin(\pi n)=\sum 0 = 0$. So, this series converges.

For any integer $n$, $\cos(\pi n)=\pm 1$, so $\lim\limits_{n \rightarrow \infty}\cos(\pi n) \neq 0$.

By the divergence test, this series diverges.

Factorials grow super fast. Like, wow, really fast. Even faster than exponentials. So the terms are going to zero, and the divergence test won't help us. Let's use ratio–it's a good go-to test with factorials.

$$\begin{aligned} \frac{a_{k+1}}{a_k}&=\frac{\frac{e^{k+1}}{(k+1)!}}{\frac{e^k}{k!}}=\frac{e^{k+1}}{e^k}\cdot\frac{k!}{(k+1)!} = e\cdot\frac{k(k-1)\cdots(1)}{(k+1)(k)(k-1)\cdots(1)}=e\cdot\frac{1}{k+1}=\frac{e}{k+1} \\ \text{Since $e$ is a constant,} \\ \lim_{k \rightarrow \infty}\frac{a_{k+1}}{a_k}&=\lim_{k \rightarrow \infty}\frac{e}{k+1}=0 \end{aligned}$$

Since $0<1$, by the ratio test, the series converges.

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 $\dfrac{1}{3}$.

Usually with factorials, we want to use the divergence test or the ratio test. Since the terms are indeed tending towards zero, we are left with the ratio test.

$$\begin{aligned} \frac{a_{n+1}}{a_n}&=\frac{\frac{(n+1)!(n+1)!}{(2n+2)!}}{\frac{n!n!}{(2n)!}}=\frac{(n+1)!(n+1)!}{n!n!}\cdot\frac{(2n)!}{(2n+2)!}\\ &=\tfrac{ (n+1){(n)}{(n-1)}\cdots{(1)}}{{n}{(n-1)}\cdots{(1)}} \cdot\tfrac{ (n+1){(n)}{(n-1)}\cdots{(1)}}{{n}{(n-1)}\cdots{(1)}} \cdot \tfrac{{(2n)}{(2n-1)}{(2n-2)}\cdots{(1)}}{ (2n+2)(2n+1){(2n)}{(2n-1)}{(2n-2)}\cdots{(1)}} \\ &=(n+1)(n+1) \cdot \frac{1}{(2n+2)(2n+1)}\text{, so}\\ \lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n}&=\lim_{n \rightarrow \infty}\frac{(n+1)(n+1)}{(2n+2)(2n+1)} =\lim_{n \rightarrow \infty}\frac{(n+1)(n+1)}{2(n+1)(2n+1)} =\lim_{n \rightarrow \infty}\frac{n+1}{4n+2}=\frac{1}{4} \end{aligned}$$

Since the limit is a number less than 1, the series converges by the ratio test.

We want to make an estimation, when $n$ gets big: $$\frac{n^2+1}{2n^4+n} \approx \frac{n^2}{2n^4}=\frac{1}{2n^2}$$ Since $\sum \frac{1}{2n^2}$ is a convergent series (by $p$-test, or integral test), we guess that our series is convergent as well. If we wanted to use comparison test, we should have to show $\frac{n^2+1}{2n^4+n}<\frac{1}{2n^2}$, which seems unpleasant, so let's use limit comparison.

$$\lim_{n \rightarrow \infty}\frac{\frac{n^2+1}{2n^4+n}}{\frac{1}{2n^2}}=\lim_{n \rightarrow \infty}\frac{(n^2+1)2n^2}{2n^4+n} =\lim_{n \rightarrow \infty}\frac{2n^4+2n^2}{2n^4+n}\left(\frac{1/n^4}{1/n^4}\right)=\lim_{n \rightarrow \infty}\frac{2+\frac{2}{n^2}}{2+\frac{1}{n^3}}=1$$

Since the limit is a positive finite number, by the Limit Comparison Test, $\sum \frac{n^1+1}{2n^4+n}$ does the same thing $\sum\frac{1}{2n^2}$ does: it converges.

As usual, let's see whether the “easy" tests work. The terms we're adding converge to zero: $$\lim_{n \to \infty}\frac{e^{-\sqrt{n}}}{\sqrt{n}}=\lim_{n \to \infty}\frac{1}{\sqrt{n}e^{\sqrt{n}}}=0$$ so the divergence test is inconclusive. Our series isn't geometric, and it doesn't seem obvious how to compare it to a geometric series. However, the terms we're adding seem like they would make an integrable function.

Set $f(x)=\frac{e^{-\sqrt{x}}}{\sqrt{x}}$. For $x\ge1$, this function is positive and decreasing (since it is the product of the two positive decreasing functions $e^{-\sqrt x}$ and $\frac1{\sqrt x}$). We use the integral test with this function. Using the substitution $u=\sqrt x$, so that $\mathrm{d}u = \frac1{2\sqrt x}\,\mathrm{d}x$, we see that

$$\begin{aligned} \int_1^\infty f(x)\,\mathrm{d}x &=\lim_{R\rightarrow\infty}\int_1^R \bigg( \frac{e^{-\sqrt{x}}}{\sqrt{x}}\,\mathrm{d}x \bigg) \\[0.05in] &=\lim_{R\rightarrow\infty} \bigg( \int_1^{\sqrt{R}} e^{-u}\cdot2\,\mathrm{d}u \bigg) \\[0.05in] &=\lim_{R\rightarrow\infty} \bigg( {-}2e^{-u}\Big|_1^{\sqrt{R}} \bigg) \\[0.05in] &=\lim_{R\rightarrow\infty} \bigg( {-}2e^{-\sqrt R} + 2e^{-\sqrt1} \bigg) = 0 + 2e^{-1}, \end{aligned}$$

and so this improper integral converges. By the integral test, the given series also converges.

We first develop some intuition. For very large $k$, $k^4$ dominates $1$ so that the numerator $\sqrt[3]{k^4+1} \approx \sqrt[3]{k^4} = k^{4/3}$, and $k^5$ dominates 9 so that the denominator $\sqrt{k^5+9} \approx \sqrt{k^5} = k^{5/2}$ and the summand

$$\begin{aligned} \frac{ \sqrt[3]{k^4+1} }{\sqrt{k^5+9}} \approx \frac{k^{4/3}}{k^{5/2}} =\frac{1}{k^{7/6}} \end{aligned}$$

The series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{k^{7/6}}$ converges by the $p$–test with $p=\frac{7}{6}>1$, so we expect the given series to converge too.

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

$$\begin{aligned} a_k= \frac{ \sqrt[3]{k^4+1} }{\sqrt{k^5+9}} \quad\text{and}\quad b_k= \frac{1}{k^{7/6}}=\frac{k^{4/3}}{k^{5/2}} \end{aligned}$$

which is valid since

$$\lim_{k\rightarrow\infty} \frac{a_k}{b_k} =\lim_{k\rightarrow\infty}\frac{ \sqrt[3]{k^4+1}/k^{4/3}} {\sqrt{k^5+9}/k^{5/2}} =\lim_{k\rightarrow\infty}\frac{ \sqrt[3]{1+1/k^4} }{\sqrt{1+9/k^5}} =1$$

exists. Since the series $\sum\limits_{k=1}^\infty b_k$ is a convergent $p$–series (with ratio $p=\frac{7}{6}>1$), the given series converges.