수열의 극한 전체 문제

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, $\lim\limits_{n \to \infty} b_n =e^3$

The $N$th term of the sequence of partial sums, $S_N$, is the sum of the first $N$ terms of the series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$. $$\begin{array}{l|l} \mathbf{N}&\mathbf{S_N}\\ \hline 1 & 1\\[7pt] 2 & 1+\frac12\\[7pt] 3 & 1+\frac12 + \frac13\\[7pt] 4 & 1+\frac12 + \frac13+\frac14\\[7pt] 5 & 1+\frac12 + \frac13+\frac14+\frac15\\[7pt] \end{array}$$

If there were a total of 17 cookies before Student 11 came, and 20 cookies after, then Student 11 brought 3 cookies.

If $f'(N)<0$, that means $f(N)$ is decreasing. So, adding more terms makes for a smaller sum. That means the terms we're adding are negative. That is, $a_n < 0$ for all $n \geq 2$.

Using logarithm rules, we see $$\displaystyle\sum_{n=5}^\infty \log\left(\frac{n-3}{n}\right)= \displaystyle\sum_{n=5}^\infty \big[\log(n-3)-\log n\big]$$ which looks like a telescoping series. Let's make a table to figure out the partial sums.

$$\begin{array}{c||cc||c} \mathbf{n}&\mathbf{\log(n-3)}&\mathbf{-\log n}&\mathbf{s_n}\\[7pt] \hline 5&\log 2&-\log 5&\log 2 - \log 5\\[7pt] 6&\log 3&-\log 6&\log 2 +\log 3- \log 5-\log 6\\[7pt] 7&\log 4&-\log 7&\log 2 +\log 3+\log 4- \log 5-\log 6 -\log 7\\[7pt] 8&\log 5&-\log 8&\log 2 +\log 3+\log 4 -\log 6 -\log 7-\log 8\\[7pt] 9&\log 6&-\log 9&\log 2 +\log 3+\log 4 -\log 7-\log 8-\log 9\\[7pt] 10&\log 7&-\log 10&\log 2 +\log 3+\log 4 -\log 8-\log 9-\log 10\\[7pt] 11&\log 8&-\log 11&\log 2 +\log 3+\log 4 -\log 9-\log 10-\log 11\\[7pt] \vdots&&& \end{array}$$

There is a “lag" before the terms cancel, which is why they “build up" more than we saw in past examples. Still, we can clearly see the $N$th partial sum:

$$\begin{aligned} \sum_{n=5}^N \Big(\log(n-3)-\log(n)\Big)&=\log 2+\log 3 + \log 4 - \log(N-2)-\log(N-1)-\log(N)\\ &=\log \left(\frac{24}{N(N-1)(N-2)}\right) \\ \text{when $N \geq 7$. So,} \\ \sum_{n=5}^\infty \Big(\log(n-3)-\log(n)\Big)&= \lim_{N \to \infty}s_N\\ &=\lim_{N \to \infty}\log \left(\frac{24}{N(N-1)(N-2)}\right) \\&=-\infty \end{aligned}$$

This is a telescoping series. Let's investigate it in the usual way. To make the pattern of cancellation clearer, we express $\frac{2}{n}-\frac{1}{n+1}-\frac{1}{n-1}= \frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n-1}$, and leave the fractions in the middle of the table unsimplified. Then every fraction has numerator one and two terms with the same denominator and opposite sign cancel.

$$\begin{array}{c||cccc||c} \mathbf{n}&\mathbf{\dfrac{1}{n}}&\mathbf{\dfrac{1}{n}}&\mathbf{-\dfrac{1}{n+1}}&\mathbf{-\dfrac{1}{n-1}}&\mathbf{s_n}\\[7pt] \hline 2&\dfrac{1}{2}&{{\dfrac{1}{2}}}&{{-\dfrac{1}{3}}}&-\dfrac{1}{1}& \dfrac{1}{2}+{{\dfrac{1}{2}}} {{-\dfrac{1}{3}}}-\dfrac{1}{1}\\[10pt] 3&{{\dfrac{1}{3}}}&{{\dfrac{1}{3}}}&{{-\dfrac{1}{4}}}&{{-\dfrac{1}{2}}}& \dfrac{1}{2}+{{\dfrac{1}{3}}}{{-\dfrac{1}{4}}}-\dfrac{1}{1}\\[10pt] 4&{{\dfrac{1}{4}}}&{{\dfrac{1}{4}}}&{{-\dfrac{1}{5}}}&{{-\dfrac{1}{3}}}& \dfrac{1}{2}+{{\dfrac{1}{4}}} {{-\dfrac{1}{5}}}-\dfrac{1}{1}\\[10pt] 5&{{\dfrac{1}{5}}}&{{\dfrac{1}{5}}}&{{-\dfrac{1}{6}}}&{{-\dfrac{1}{4}}}& \dfrac{1}{2}+{{\dfrac{1}{5}}}{{-\dfrac{1}{6}}}-\dfrac{1}{1}\\[10pt] 6&{{\dfrac{1}{6}}}&{{\dfrac{1}{6}}}&{{-\dfrac{1}{7}}}&{{-\dfrac{1}{5}}}& \dfrac{1}{2}+{{\dfrac{1}{6}}}{{-\dfrac{1}{7}}}-\dfrac{1}{1}\\[10pt] 7&{{\dfrac{1}{7}}}&{{\dfrac{1}{7}}}&{{-\dfrac{1}{8}}}&{{-\dfrac{1}{6}}}& \dfrac{1}{2}+{{\dfrac{1}{7}}} {{-\dfrac{1}{8}}}-\dfrac{1}{1}\\[10pt] 8&{{\dfrac{1}{8}}}&\dfrac{1}{8}&-\dfrac{1}{9}&{{-\dfrac{1}{7}}}& \dfrac{1}{2}+\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{1}\\[10pt] \vdots&&&& \end{array}$$

Concentrate on any row $n$, except the very first row and the very last row. The first $\frac{1}{n}$ in that row cancels the $-\frac{1}{n}$ in the middle of the row above it, and the second $\frac{1}{n}$ in that row cancels the $-\frac{1}{n}$ at the end of the row below it. As far as the first ($n=2$) row is concerned, the first $\frac{1}{2}$ and the last $-\frac{1}{1}$ never get cancelled out because there is no row above the first one. And as far as the very last row is concerned, the two middle terms never get cancelled out because there is no row after the last one. So the partial sum is

$$\begin{aligned} s_N&=\sum_{n=2}^N \left(\frac{2}{n}-\frac{1}{n+1}-\frac{1}{n-1} \right) =\overbrace{\frac{1}{2}-\frac{1}{1}}^{\text{from the first row}} +\overbrace{\frac{1}{N}-\frac{1}{N+1}}^{\text{from the last row}} \end{aligned}$$

There is another purely algebraic way to find the same $s_N$, motivated by the above discussion.

$$\begin{aligned} s_N&=\sum_{n=2}^N \left(\frac{2}{n}-\frac{1}{n+1}-\frac{1}{n-1} \right) \\ &=\sum_{n=2}^N \left(\frac{1}{n}-\frac{1}{n+1}\right) +\sum_{n=2}^N\left(\frac{1}{n}-\frac{1}{n-1} \right) \end{aligned}$$

The first half

$$\begin{aligned} \sum_{n=2}^N \left(\frac{1}{n}-\frac{1}{n+1}\right) &=\left(\frac{1}{2}{{-\frac{1}{3}}}\right) + \left({{\frac{1}{3}}} {{-\frac{1}{4}}}\right) + \left({{\frac{1}{4}}}{{-\frac{1}{5}}}\right) + \cdots +\left({{\frac{1}{N}}}-\frac{1}{N+1}\right) \\ &=\left[\frac{1}{2}-\frac{1}{N+1}\right] \end{aligned}$$

and the second half

$$\begin{aligned} \sum_{n=2}^N\left(\frac{1}{n}-\frac{1}{n-1} \right) &=\left({{\frac{1}{2}}}-\frac{1}{1} \right) + \left({{\frac{1}{3}}}{{-\frac{1}{2}}} \right) + \left({{\frac{1}{4}}}{{-\frac{1}{3}}}\right) +\cdots + \left(\frac{1}{N}-{{\frac{1}{N-1}}} \right) \\ &=\left\{-\frac{1}{1}+\frac{1}{N}\right\} \end{aligned}$$

So

$$\begin{aligned} s_N&=\left[\frac{1}{2}-\frac{1}{N+1}\right]+\left\{-\frac{1}{1}+\frac{1}{N}\right\} \\ &=-\frac{1}{2}+\frac{1}{N}-\frac{1}{N+1} \end{aligned}$$

and the limit

$$\begin{aligned} \sum_{n=2}^\infty \left(\frac{2}{n}-\frac{1}{n+1}-\frac{1}{n-1} \right)&=\lim_{N \to \infty}s_N\\&=\lim_{N \to \infty} \left[-\frac{1}{2} +\frac{1}{N}-\frac{1}{N+1}\right]\\ &=-\frac{1}{2} \end{aligned}$$