비교 판정법

${S}_{10}\approx 383,$ $\lim_{n\to \infty }{S}_{n}=400$

(a) If Olaf is old, and I am even older, then I am old as well.

(b) If Olaf is old, and I am not as old, then perhaps I am old as well (just slightly less so), or perhaps I am young. There is not enough information to tell.

(c) If Yuan is young, and I am older, then perhaps I am much older and I am old, or perhaps I am only a little older, and I am young. There is not enough information to tell.

(d) If Yuan is young, and I am even younger, then I must also be young.

Another way to think about this is with a timeline of birthdates. People born before the threshold are old, and people born after it are young.

If I'm born before (older than) Olaf, I'm born before the threshold, so I'm old.

If I'm born after (younger than) Yuan I'm born after the threshold, so I'm young.

If I'm born after Olaf or before Yuan, I don't know which side of the threshold I'm on. I could be old or I could be young.

The integral test does not apply because $f(x)$ is not decreasing.

One possible answer: $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}$. This series converges (it's a $p$–series with $p=2>1$), but if we take the ratio of consecutive terms: $$\lim_{n \to \infty}\frac{a_{n+1}}{a_n} = \lim_{n \to \infty}\frac{n^2}{(n+1)^2}=1$$ The limit of the ratio is 1, so the ratio test is inconclusive.

By the divergence test, for a series $\sum a_n$ to converge, we need $\lim\limits_{n \to \infty} a_n=0$. That is, the magnitude (absolute value) of the terms needs to be getting smaller. If $\displaystyle\lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right|<1$ or (equivalently) $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|>1$, then $|a_{n+1}|>|a_n|$ for sufficiently large $n$, so the terms are actually growing in magnitude. That means the series diverges, by the divergence test.

The terms of the series only see a small portion of the domain of the integral. We can try to think of a function $f(x)$ that behaves “nicely" when $x$ is a whole number (that is, it produces a sequence whose sum converges), but is more unruly when $x$ is not a whole number.

For example, suppose $f(x)=\sin(\pi x)$. Then $f(x)=0$ for every integer $x$, but this is not representative of the function as a whole. Indeed, our corresponding series has terms $\{a_n\}=\{0,0,0,\ldots\}$.

• $\displaystyle\int_{1}^\infty \sin(\pi x)\,\mathrm{d}x = \lim_{R \to \infty} \left[-\frac{1}{\pi}\cos (\pi x)\right]_1^R = \lim_{R \to \infty}\Big[-\cos (\pi R)\Big]-\frac{1}{\pi}$

Since the limit does not exist, the integral diverges.

• $\displaystyle\sum_{n=1}^{\infty}\sin(\pi n) = \sum_{n=1}^\infty 0 = 0$. The series converges.

(a) In general false. The harmonic series $\sum\limits_{n=1}^\infty\frac{1}{n}$ diverges by the $p$–test with $p=1$.

(b) Be careful. You were not told that the $a_n$'s are positive. So this is false in general. If $a_n =(-1)^n\frac{1}{n}$, then $\sum\limits_{n=1}^{\infty} (-1)^{n\mathstrut} a_n$ is again the harmonic series $\sum\limits_{n=1}^\infty\frac{1}{n}$, which diverges.

(c) In general false. Take, for example, $a_n=0$ and $b_n=1$.

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.