수열의 극한 전체 문제

$\sum_{n=1}^{\infty }{a}_{n}=f(1)-f(2)$

Let's take stock: $\sin(1/n) \to \sin (0)=0$ as $n \to \infty$, so $\log\left(\sin(1/n)\right) \to -\infty$. However, $\log(2n) \to \infty$. So, we have some tension here: the two pieces behave in ways that pull the terms of the sequence in different directions. (Recall we cannot conclude anything like “$-\infty+\infty=0$.")

We try using logarithm rules to get a clearer picture.

$$\begin{aligned} \log\left(\sin\frac{1}{n}\right)+\log(2n)&=\log\left(2n\sin\left(\frac1n\right)\right) \end{aligned}$$

Still, we have indeterminate behaviour: $2n\sin(1/n)$ is the product of $2n$, which grows without bound, and $\sin(1/n)$, which approaches zero. In the past, we learned that we can handle the indeterminate form $0\cdot\infty$ with l'Hopital's rule (after a little algebra), but there's a slicker way. Note $1/n \to 0$ as $n \to \infty$. If we write $\frac{1}{n}=x$, then this piece of our limit resembles something familiar.

$$\begin{aligned} 2n\sin\left(\frac{1}{n}\right)&=2\left(\frac{\sin x}{x}\right) \\ \text{If $n \to \infty$, then $x = \frac{1}{n}\to 0$.} \\ \lim_{n \to \infty}2n\sin\left(\frac{1}{n}\right)&=2\lim_{x \to 0}\frac{\sin x}{x} \\ \text{That limit is familiar:} \\ &=2(1)=2 \\ \text{Then:} \\ \lim_{n \to \infty}\log\left(2n\sin\left(\frac{1}{n}\right)\right)&=\log 2 \end{aligned}$$

Note: if you have forgotten that $\displaystyle\lim_{x \to 0}\frac{\sin x}{x}=1$, you can also evaluate this limit using l'Hopital's rule:

$$\begin{aligned} \underbrace{\lim_{x \to 0}\frac{\sin x}{x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to 0}{\mathrm{den}\to0}} &=\lim_{x \to 0}\frac{\cos x}{1}=\cos 0 =1 \end{aligned}$$

First, although this sequence is not defined for some small values of $n$, it is defined as long as $n \geq 5$, so it's not a problem to take the limit as $n \to \infty$. Second, we notice that our limit has the indeterminate form $\infty-\infty$. Since this form is indeterminate, more work is needed to find our limit, if it exists.

A standard trick we saw last semester with functions of this form was to multiply and divide by the conjugate of the expression, $\sqrt{n^2+5n}+\sqrt{n^2-5n}$. Then the denominator will be the sum of two similar things, rather than their difference. See the work below to find out why that is helpful.

$$\begin{aligned} \sqrt{n^2+5n}-\sqrt{n^2-5n}&=\big(\sqrt{n^2+5n}-\sqrt{n^2-5n}\big)\left(\frac{\sqrt{n^2+5n}+\sqrt{n^2-5n}}{\sqrt{n^2+5n}+\sqrt{n^2-5n}}\right)\\ &=\frac{(n^2+5n)-(n^2-5n)}{\sqrt{n^2+5n}+\sqrt{n^2-5n}}\\ &=\frac{10n}{\sqrt{n^2+5n}+\sqrt{n^2-5n}} \\ \text{Now, we'll cancel out $n$ from the top and the bottom. Note $n=\sqrt{n^2}$.} \\ &=\frac{10n}{\sqrt{n^2+5n}+\sqrt{n^2-5n}}\left(\frac{\frac1n}{\frac1n}\right)\\ &=\frac{10n}{\sqrt{n^2+5n}+\sqrt{n^2-5n}}\left(\frac{\frac1n}{\frac{1}{\sqrt{n^2}}}\right)\\ &=\frac{10}{\sqrt{1+\frac5n}+\sqrt{1-\frac5n}} \\ \text{Now, the limit is clear.} \\ \lim_{n \to \infty}\frac{10}{\sqrt{1+\frac5n}+\sqrt{1-\frac5n}}&=\frac{10}{\sqrt{1+0}+\sqrt{1+0}}=\frac{10}{1+1}=5 \end{aligned}$$

First, we note that we have in indeterminate form: as $n$ grows, $2+\frac1n \to 2$, so $n\left[\left(2+\frac1n\right)^{100}-2^{100}\right]$ has the form $\infty \cdot 0$. To overcome this difficulty, we could use some algebra and l'Hopital's rule, but there's a slicker way. If we let $h = \frac{1}{n}$, then $h \to 0$ as $n\to\infty$, and our limit looks like:

$$\begin{aligned} \lim_{n \to \infty}n\left[\left(2+\frac1n\right)^{100}-2^{100}\right]&= \lim_{h \to 0}\frac{\left(2+h\right)^{100}-2^{100}}{h} \\ \text{This reminds us of the definition of a derivative.} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{x^{100}\right\}&=\lim_{h \to 0}\frac{(x+h)^{100}-x^{100}}{h} \end{aligned}$$

So, if we set $f(x)=x^{100}$, our limit is simply $f'(2)$. That is, $\left[100x^{99}\right]_{x=2}= 100\cdot 2^{99}$.

If we naively try to find the limit, we run up against the indeterminate form $1^{\infty}$. We'd like to use l'Hopital's rule, but we don't have the form $\frac{\infty}{\infty}$ or $\frac{0}{0}$–we'll need to use a logarithm. Additionally, l'Hopital's rule applies to differentiable functions defined for real numbers–so we'll consider a function, rather than the sequence.

Note the terms of the sequence are all positive.

• Solution 1: Define $x=\frac{1}{n}$, and $f(x)=\left(1+3x+5x^2\right)^{1/x}$. Then $b_n=f\left(\frac{1}{n}\right)=f(x)$, and $$\lim_{n \to \infty} f\left(\frac{1}{n}\right) = \lim_{x \to 0^+}f(x).$$ If this limit exists, it is equal to $\lim\limits_{n \to \infty}b_n$.

$$\begin{aligned} \lim_{x \to 0^+}f(x)&= \lim_{x \to 0^+} \left(1+3x+5x^2\right)^{1/x}\\ \lim_{x \to 0^+}\log[f(x)]&= \lim_{x \to 0^+} \log\left[\left(1+3x+5x^2\right)^{1/x}\right] =\underbrace{ \lim_{x \to 0^+}\frac{ \log\left[1+3x+5x^2 \right]}{x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to 0}{\mathrm{den}\to0}}\\ &=\lim_{x \to 0^+}\frac{\frac{3+10x}{1+3x+5x^2}}{1}=3\\ \lim_{x \to 0^+}f(x)&=e^3 \end{aligned}$$

Since the limit exists, $\lim\limits_{n \to \infty}b_n=e^3$.

• Solution 2: If we didn't see the nice simplifying trick of letting $x=\frac{1}{n}$, we can still solve the problem using $g(x)=\left(1+\frac{3}{x}+\frac{5}{x^2}\right)^{x}$:

$$\begin{aligned} g(x)&=\left(1+\frac{3}{x}+\frac{5}{x^2}\right)^x\\ \log [g(x)]&= x\log\left[1+\frac{3}{x}+\frac{5}{x^2}\right] =\underbrace{\frac{\log\left[1+\frac{3}{x}+\frac{5}{x^2}\right]}{1/x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to 0}{\mathrm{den}\to0}}\\ \lim_{x \to \infty} \log[g(x)]&=\lim_{x \to \infty}\dfrac{\frac{-\frac{3}{x^2}-\frac{10}{x^3}}{1+\frac{3}{x}+\frac{5}{x^2}}}{\frac{-1}{x^2}}= \lim_{x \to \infty}x^2\frac{\frac{3}{x^2}+\frac{10}{x^3}}{1+\frac{3}{x}+\frac{5}{x^2}} = \lim_{x \to \infty}\frac{3+\frac{10}{x}}{1+\frac{3}{x}+\frac{5}{x^2}} = \frac{3+0}{1+0+0}=3\\ \lim_{x \to \infty} f(x)&=e^3\end{aligned}$$

Since the limit exists, $\lim\limits_{n \to \infty} b_n =e^3$