國立臺灣大學一百學年度研究所碩士班入學考試試題：微積分乙（不含線性代數）

1. ($30\%$) Suppose a random value $y$ follows the $g$ distribution, i.e., $y$ can be expressed as a function of a normally distributed variable as follows.

   $\displaystyle y=\delta+\lambda\frac{\exp\left(gz\right)-1}g$,

   where $\delta$ and $g$ are real numbers, $\lambda$ is a positive real number, and $z$ follows the normal distribution with mean to be $\mu$ and standard deviation to be $\sigma$. Given the constraint $\left[g\left(y-\delta\right)/\lambda+1\right]>0$, solve the following problems.
   1. ($4\%$) Express $z$ in terms of $y$, i.e., find the function $h\left(y\right)=z$.
   2. ($4\%$) With the function $h\left(y\right)$, derive the density function of $y$, $f\left(y\right)$, according to the definition $\displaystyle f\left(y\right)=\frac1{\sigma\sqrt{2\pi}}\left|h'\left(y\right)\right|e^{-\frac1{2\sigma^2}\left(h\left(y\right)-\mu\right)^2}$, where $\mu$ and $\sigma$ are the mean and the standard deviation of $z$.
   3. ($8\%$) Derive the expectation of $y$, $E(y)$.
   4. ($8\%$) Derive the variance of $y$, $\text{var}(y)$.
   5. ($6\%$) Show that when $g$ is zero, $y$ follows a normal distribution. [Hint: Consider the result of $\displaystyle\lim_{g\to0}\frac{\exp\left(gz\right)-1}g$.]
訣竅
前兩小題根據題目給定的資訊整理即可，而後三小題則分別按照期望值、變異數與常態分配的定義處理。
解法
1. 移項便有

   $\displaystyle \frac{g\left(y-\delta\right)}{\lambda}+1=\exp\left(gz\right)$

   由於 $\left[g\left(y-\delta\right)\lambda+1\right]>0$，故可同取自然對數便有 $\displaystyle gz=\ln\left(1+\frac{g\left(y-\delta\right)}{\lambda}\right)$，故得

   $\displaystyle z=\frac1g\ln\left(1+\frac{g\left(y-\delta\right)}{\lambda}\right):=h\left(y\right)$

2. 先求 $h$ 的導函數有

   $\displaystyle h'\left(y\right)=\frac1g\cdot\frac{g}{\lambda+g\left(y-\delta\right)}=\frac1{\lambda+g\left(y-\delta\right)}$

   由於 $g\left(y-\delta\right)/\lambda+1>0$，故 $\lambda+g\left(y-\delta\right)>0$，故 $\left|h'\left(y\right)\right|=h'\left(y\right)$。從而 $y$ 的機率密度函數為

   $\displaystyle f\left(y\right)=\frac1{\sigma\sqrt{2\pi}}\cdot\frac1{\lambda+g\left(y-\delta\right)}\exp\left[-\frac1{2\sigma^2}\left(\frac{\ln\left(\lambda+g\left(y-\delta\right)\right)-\ln\lambda}g-\mu\right)^2\right]$

   其中 $y\in\left(y^*,\infty\right)$，$y^*=-\lambda/g+\delta$。
3. 那麼由期望值的定義可知

   $\displaystyle E\left(y\right)=\int_{y^*}^{\infty}yf\left(y\right)dy$

   令 $\displaystyle t=\frac{h\left(y\right)-\mu}{\sigma\sqrt2}$，那
   • 當 $y\to y^*$ 時有 $t\to-\infty$；
   • 當 $y\to\infty$ 時有 $t\to\infty$；
   • 求導有 $\displaystyle dt=\frac{h'\left(y\right)}{\sigma\sqrt2}dy$。再者整理有 $y=h^{-1}\left(\mu+t\sigma\sqrt2\right)$。
   如此所求為

   $\displaystyle\begin{aligned}E\left(y\right)&=\frac1{\sqrt\pi}\int_{-\infty}^{\infty}h^{-1}\left(\mu+t\sigma\sqrt2\right)e^{-t^2}dt=\frac1{\sqrt\pi}\int_{-\infty}^{\infty}\left[\delta-\frac{\lambda}g+\frac{\lambda}g\exp\left(g\mu+tg\sigma\sqrt2\right)\right]e^{-t^2}dt\\&=\frac1{\sqrt\pi}\int_{-\infty}^{\infty}\left(\delta-\frac{\lambda}g\right)e^{-t^2}dt+\frac{\lambda e^{g\mu}}{g\sqrt\pi}\int_{-\infty}^{\infty}\exp\left(-t^2+tg\sigma\sqrt2\right)dt\\&=\delta-\frac{\lambda}g+\frac{\lambda e^{g\mu+g^2\sigma^2/2}}{g}\end{aligned}$

4. 按照變異數的定義有 $\text{var}\left(y\right)=E\left(y^2\right)-\left(E\left(y\right)\right)^2$，故我們僅需計算 $E\left(y^2\right)$，即計算

   $\displaystyle E\left(y^2\right)=\int_{y^*}^{\infty}y^2f\left(y\right)dy$

   沿用前一小題的代換可知

   $\displaystyle\begin{aligned}E\left(y^2\right)&=\frac1{\sqrt\pi}\int_{-\infty}^{\infty}\left[\delta-\frac{\lambda}g+\frac{\lambda}g\exp\left(g\mu+tg\sigma\sqrt2\right)\right]^2e^{-t^2}dt\\&=\frac1{\sqrt\pi}\int_{-\infty}^{\infty}\left(\delta-\frac{\lambda}g\right)^2e^{-t^2}dt+\frac{2\lambda e^{g\mu}}{g\sqrt\pi}\left(\delta-\frac{\lambda}g\right)\int_{-\infty}^{\infty}\exp\left(-t^2+tg\sigma\sqrt2\right)dt+\frac{\lambda^2e^{2g\mu}}{g^2\sqrt\pi}\int_{-\infty}^{\infty}\exp\left(-t^2+2tg\sigma\sqrt2\right)dt\\&=\left(\delta-\frac{\lambda}g\right)^2+\frac{2\lambda e^{g\mu+g^2\sigma^2/2}}g\left(\delta-\frac{\lambda}g\right)+\frac{\lambda^2e^{2g\mu+2g^2\sigma^2}}{g^2}\end{aligned}$

   因此所求為

   $\displaystyle\begin{aligned}\text{var}\left(y\right)&=\left[\left(\delta-\frac{\lambda}g\right)^2+\frac{2\lambda e^{g\mu+g^2\sigma^2/2}}g\left(\delta-\frac{\lambda}g\right)+\frac{\lambda^2e^{2g\mu+2g^2\sigma^2}}{g^2}\right]-\left(\delta-\frac{\lambda}g+\frac{\lambda e^{g\mu+g^2\sigma^2/2}}g\right)^2\\&=\frac{\lambda^2}{g^2}e^{2g\mu+g^2\sigma^2}\left(e^{g^2\sigma^2}-1\right)\end{aligned}$

5. 由於當 $g$ 趨於零時可以發現 $\displaystyle\lim_{g\to0}\frac{\exp\left(gz\right)-1}g=\lim_{g\to0}\frac{z\exp\left(gz\right)}1=z$。故當 $g=0$ 時有 $y=\delta+\lambda z$，此表明經由線性轉換能使 $y$ 成為平均為 $\delta+\lambda\mu$，而標準差則為 $\lambda\sigma$。由 (c) 與 (d) 能知 $\displaystyle\lim_{g\to0}E\left(y\right)=\delta+\lambda\mu$ 且 $\displaystyle\lim_{g\to0}\text{var}\left(y\right)=\lambda^2\sigma^2$。


2. ($8\%$) Find the maximum and minimum values of $f\left(x,y,z\right)=x+3y-z$ subject to $4x^2+2y^2+z^2=4$.
訣竅
運用初等不等式求極值；亦可使用拉格朗日乘子法求解。
解法一
由柯西不等式可知
$\displaystyle23=\left[\left(2x\right)^2+\left(\sqrt2y\right)^2+z^2\right]\left[\left(\frac12\right)^2+\left(\frac3{\sqrt2}\right)^2+\left(-1\right)^2\right]\geq\left(x+3y-z\right)^2$
因此函數 $f$ 在給定的限制條件下的最大最小值分別為 $\sqrt{23}$ 與 $-\sqrt{23}$，此時的等號成立條件為 $\displaystyle 4x=\frac{2y}3=-z$，即 $\displaystyle\left(\frac1{\sqrt{23}},\frac6{\sqrt{23}},-\frac4{\sqrt{23}}\right)$ 與 $\displaystyle\left(-\frac1{\sqrt{23}},-\frac6{\sqrt{23}},\frac4{\sqrt{23}}\right)$。
解法二
設定拉格朗日乘子函數如下
$\displaystyle F\left(x,y,z,\lambda\right)=x+3y-z+\lambda\left(4x^2+2y^2+z^2-4\right)$
據此解下列的聯立方程組
$\left\{\begin{aligned}&F_x\left(x,y,z,\lambda\right)=1+8\lambda x=0\\&F_y\left(x,y,z,\lambda\right)=3+4\lambda y=0\\&F_z\left(x,y,z,\lambda\right)=-1+2\lambda z=0\\&F_{\lambda}\left(x,y,z,\lambda\right)=4x^2+2y^2+z^2-4=0\end{aligned}\right.$
由於明顯 $\lambda\neq0$，故 $\displaystyle x=-\frac1{8\lambda}$、$\displaystyle y=-\frac3{4\lambda}$、$\displaystyle z=\frac1{2\lambda}$，由此代入第四式為
$\displaystyle\frac4{64\lambda^2}+\frac{2\cdot9}{16\lambda^2}+\frac1{4\lambda^2}=4$
即得 $\lambda^2=\frac{23}{64}$，故有 $\displaystyle\lambda=\pm\frac{\sqrt{23}}8$。如此得兩座標
$\displaystyle\left(x,y,z\right)=\frac{\left(1,6,-4\right)}{\sqrt{23}}$,　$\displaystyle\left(x,y,z\right)=\frac{\left(-1,-6,4\right)}{\sqrt{23}}$
而最大最小值分別為 $\sqrt{23}$ 與 $-\sqrt{23}$。


3. ($12\%$) A $10,000$-cubic-foot-room has $500$ smoke particles per cubic foot. A ventilation system is turned on that each minute brings in $500$ cubic feet of smoke-free air, while an equal volume of air leaves the room. Also, during each minute, smoker in the room add a total of $10,000$ particles of smoke to the room. Assume that the air in the room mixes thoroughly.
   1. ($4\%$) Find a differential equation and initial condition that govern the total number $y\left(t\right)$ of smoke particles in the room after $t$ minutes.
   2. ($4\%$) Solve this differential equation.
   3. ($4\%$) Find how soon the smoke level will fall to $100$ smoke particles per cubic foot. (If the answers are with decimal numbers, please round to the nearest hundredth.)
訣竅
按照濃度的變化率寫出出微分方程並求解。
解法
1. 由於 $y\left(t\right)$ 表在時刻 $t$ 分鐘後的煙霧粒子的個數，那麼 $y'\left(t\right)$ 便表示在 $t$ 分鐘後增加煙霧粒子的瞬間增加量，按照題意可知

   $\displaystyle y'\left(t\right)=10000-\frac{y\left(t\right)}{10000}\cdot500$

   再者一開始房間內有 $500\cdot10^4$ 個煙霧粒子，因此 $y\left(0\right)=5\cdot10^6$。
2. 經由移項可知

   $\displaystyle\frac{dy}{200000-y}=\frac{dt}{20}$

   同取積分有

   $\displaystyle-\ln\left|200000-y\right|=\frac{t}{20}+C$

   透過初始條件可知 $C=-\ln4800000$，從而所求為

   $y\left(t\right)=200000+4800000e^{-t/20}$

3. 要達到濃度為每立方英尺有 $100$ 顆煙霧粒子，則該房間要有總數 $10^6$ 顆煙霧粒子。因此解方程式 $y\left(t\right)=10^6$，即 $2+48e^{-t/20}=10$，故有 $t=20\ln6$。


4. Evaluate:
   1. ($5\%$) $\displaystyle\phi\left(\alpha\right)=\int_0^{\infty}\alpha e^{-\alpha x}dx$ for $\alpha>0$.
   2. ($10\%$) Evaluate $\displaystyle\int_0^1e^{x^2}dx$, approximately by using Taylor's theorem of the mean and estimate the maximum error.
訣竅
直接演算瑕積分並透過帶餘項的泰勒定理估算瑕積分並給出誤差估計。
解法
1. 直接計算瑕積分可知

   $\displaystyle\phi\left(\alpha\right)=-e^{-\alpha x}\Big|_0^{\infty}=1$

2. 回憶自然指數的泰勒展開式為

   $\displaystyle e^x=\sum_{k=0}^n\frac{x^k}{k!}+\frac{e^{\xi}}{\left(n+1\right)!}x^{n+1}$

   此處 $\xi$ 介於 $0$ 與 $x$ 之間。現用 $x^2$ 取代 $x$ 有

   $\displaystyle e^{x^2}=\sum_{k=0}^n\frac{x^{2k}}{k!}+\frac{e^{\eta^2}}{\left(n+1\right)!}x^{2n+2}$

   而此時的 $\eta$ 介於 $0$ 與 $x$ 之間。由此計算定積分可知其近似值為

   $\displaystyle\int_0^1e^{x^2}dx\approx\int_0^1\sum_{k=0}^n\frac{x^{2k}}{k!}dx=\sum_{k=0}^n\frac1{k!\left(2k+1\right)}$

   而其最大的誤差估計為

   $\displaystyle\left|\int_0^1e^{x^2}dx-\sum_{k=0}^n\frac1{k!\left(2k+1\right)}\right|\leq\int_0^1\frac{e^{\eta^2}x^{2n+2}}{\left(n+1\right)!}dx=\frac{e^{\eta^2}}{\left(n+1\right)!\left(2n+3\right)}\leq\frac{e}{\left(n+1\right)!\left(2n+3\right)}$

   這是因為 $\eta\in\left[0,x\right]\subseteq\left[0,1\right]$，因此 $e^{\eta^2}\leq e$。


5. Let $\varphi$ be a continuous real function on $\left[0,\alpha\right]$. Assume that $\varphi\left(x\right)>0$ if $0<x\leq\alpha$ and that $\varphi\left(x\right)\sim Ax^r$ as $x\to0$ ($A>0$, $r\geq0$). For all $t>0$ set

   $\displaystyle F\left(t\right)=\int_0^{\alpha}\frac{dx}{t+\varphi\left(x\right)}$.

   1. ($10\%$) Show that if $r>1$, then as $t\to0$

      $\displaystyle F\left(t\right)\sim\frac{\pi}{rA^{1/r}\sin\left(\pi/r\right)}\cdot\frac1{t^{1-1/r}}$.

   2. ($10\%$) Show that if $r=1$, then as $t\to0$

      $\displaystyle F\left(t\right)\sim\frac1A\log\frac1t$.

訣竅
由於題目所提供的資訊為在原點附近的漸近行為，故適當的將給定的積分範圍分割後以便計算。
解法

1. 給定正數 $\varepsilon<1$，那麼存在 $\delta>0$使得 $0<x<\delta$ 蘊含 $Ax^r\left(1-\varepsilon\right)<\varphi\left(x\right)<Ax^r\left(1+\varepsilon\right)$。據此我們可寫 $F$ 如下

   $\displaystyle F\left(t\right)=\int_0^{\delta}\frac{dx}{t+\varphi\left(x\right)}+\int_{\delta}^{\alpha}\frac{dx}{t+\varphi\left(x\right)}$

   對於第二項可得如下的估算

   $\displaystyle0<\int_{\delta}^{\alpha}\frac{dx}{t+\varphi\left(x\right)}\leq\int_{\delta}^{\alpha}\frac{dx}{\displaystyle\min_{\left[\delta,\alpha\right]}\varphi}=\frac{\alpha-\delta}{\displaystyle\min_{\left[\delta,\alpha\right]}\varphi}<\infty$

   故有 $t_1^*$ 使得 $0<t<t_1^*$ 蘊含

   $\displaystyle\int_{\delta}^{\alpha}\frac1{t+\varphi\left(x\right)}<\frac{\varepsilon}{2rA^{1/r}\left(1+\varepsilon\right)^{1/r}t^{1-1/r}}$

   至於第一項則容易知道

   $\displaystyle\frac1t\int_0^{\delta}\frac{dx}{1+Ax^r\left(1+\varepsilon\right)/t}=\int_0^{\delta}\frac{dx}{t+Ax^r\left(1+\varepsilon\right)}<\int_0^{\delta}\frac{dx}{t+\varphi\left(x\right)}<\int_0^{\delta}\frac{dx}{t+Ax^r\left(1-\varepsilon\right)}=\frac1t\int_0^{\delta}\frac{dx}{1+Ax^r\left(1-\varepsilon\right)/t}$

   首先我們處理積分 $\displaystyle\int_0^{\delta}\frac{dx}{1+Ax^r\left(1+\varepsilon\right)/t}$，令 $y=1+Ax^r\left(1+\varepsilon\right)/t$，那麼有
   • 當 $x=0$ 時有 $y=1$；
   • 當 $x=\delta$ 時有 $y=1+A\delta^r\left(1+\varepsilon\right)/t$；
   • 整理有 $\displaystyle x=\left(\frac{t\left(y-1\right)}{A\left(1+\varepsilon\right)}\right)^{1/r}$，求導有 $\displaystyle dx=\frac1r\left(\frac{t}{A\left(1+\varepsilon\right)}\right)^{1/r}\left(y-1\right)^{\left(1-r\right)/r}dy$。
   如此可知

   $\displaystyle\begin{aligned}\int_0^{\delta}\frac{dx}{1+Ax^r\left(1+\varepsilon\right)/t}&=\frac{t^{1/r}}{rA^{1/r}\left(1+\varepsilon\right)^{1/r}}\int_1^{1+A\delta^r\left(1+\varepsilon\right)/t}\frac1{y\left(y-1\right)^{1-1/r}}dy\\&=\frac{t^{1/r}}{rA^{1/r}\left(1+\varepsilon\right)^{1/r}}\left[\int_1^{\infty}\frac1{y\left(y-1\right)^{1-1/r}}dy-\int_{1+A\delta^r\left(1+\varepsilon\right)/t}^{\infty}\frac1{y\left(y-1\right)^{1-1/r}}dy\right]\end{aligned}$

   由於 $\displaystyle\lim_{y\to\infty}\frac{\displaystyle\frac1{y\left(y-1\right)^{1-1/r}}}{\displaystyle\frac1y}=0$，故瑕積分 $\displaystyle\int_{1+A\delta^r\left(1+\varepsilon\right)/t}^{\infty}\frac1{y\left(y-1\right)^{1-1/r}}dy$ 有限，從而 $\displaystyle\lim_{t\to0^+}\int_{1+A\delta^r\left(1+\varepsilon\right)/t}^{\infty}\frac1{y\left(y-1\right)^{1-1/r}}dy=0$，即當 $t<t_2^*$ 時有

   $\displaystyle\int_{1+A\delta^r\left(1+\varepsilon\right)/t}^{\infty}\frac1{y\left(y-1\right)^{1-1/r}}dy<\frac\varepsilon2$

   另一方面，透過變換 $u=y-1$ 可知

   $\displaystyle\int_1^{\infty}\frac1{y\left(y-1\right)^{1-1/r}}dy=\int_0^{\infty}\frac{u^{1/r-1}}{1+u}du$

   運用複變函數論中的留數定理與路徑積分（可參見「複變數－原理及題解」，中央圖書出版社，吳新生編著，第 403 頁）可知

   $\displaystyle\int_0^{\infty}\frac{u^{1/r-1}}{1+u}du=\frac{\pi}{\sin\left(\pi/r\right)}$

   綜合上述的推導可得

   $\displaystyle\frac1{rA^{1/r}\left(1+\varepsilon\right)^{1/r}}\left(\frac{\pi}{\sin\left(\pi/r\right)}-\varepsilon\right)\frac1{t^{1-1/r}}<F\left(t\right)$

   類似地，我們也有

   $\displaystyle F\left(t\right)<\frac1{rA^{1/r}\left(1-\varepsilon\right)^{1/r}}\left(\frac{\pi}{\sin\left(\pi/r\right)}+\varepsilon\right)\frac1{t^{1-1/r}}$

   將兩者結合可知

   $\displaystyle\frac1{rA^{1/r}\left(1+\varepsilon\right)^{1/r}}\left(\frac{\pi}{\sin\left(\pi/r\right)}-\varepsilon\right)<t^{1-1/r}F\left(t\right)<\frac1{rA^{1/r}\left(1-\varepsilon\right)^{1/r}}\left(\frac{\pi}{\sin\left(\pi/r\right)}+\varepsilon\right)$

   故取由 $t\to0^+$ 與 $\varepsilon$ 的任意性可知

   $\displaystyle\lim_{t\to0^+}t^{1-1/r}F\left(t\right)=\frac\pi{rA^{1/r}\sin\left(\pi/r\right)}$

   這就給出了

   $\displaystyle F\left(t\right)\sim\frac\pi{rA^{1/r}\sin\left(\pi/r\right)}\frac1{t^{1-1/r}}$

2. 承前述步驟可知

   $\displaystyle\int_0^{\delta}\frac{dx}{t+Ax\left(1+\varepsilon\right)}<\int_0^{\delta}\frac{dx}{t+\varphi\left(x\right)}<\int_0^{\delta}\frac{dx}{t+Ax\left(1-\varepsilon\right)}$

   直接計算可給出

   $\displaystyle\frac1{A\left(1+\varepsilon\right)}\log\left(1+\frac{A\delta\left(1+\varepsilon\right)}t\right)<\int_0^{\delta}\frac{dx}{t+\varphi\left(x\right)}<\frac1{A\left(1-\varepsilon\right)}\log\left(1+\frac{A\delta\left(1-\varepsilon\right)}t\right)$

   同乘以 $\displaystyle\frac{A}{\log\left(1/t\right)}$ 有

   $\displaystyle\frac1{1+\varepsilon}\frac{\log\left(1+A\delta\left(1+\varepsilon\right)/t\right)}{\log\left(1/t\right)}<\frac{A}{\log\left(1/t\right)}\int_0^{\delta}\frac{dx}{t+\varphi\left(x\right)}<\frac1{1-\varepsilon}\frac{\log\left(1+A\delta\left(1-\varepsilon\right)/t\right)}{\log\left(1/t\right)}$

   取 $t$ 趨於零且由 $\varepsilon$ 的任意性可知 $\displaystyle\lim_{t\to0^+}\frac{A}{\log\left(1/t\right)}\int_0^{\delta}\frac{dx}{t+\varphi\left(x\right)}=1$，從而有

   $\displaystyle\lim_{t\to0^+}\frac{A}{\log\left(1/t\right)}F\left(t\right)=1$

   這就給出了

   $\displaystyle F\left(t\right)\sim\frac1A\log\frac1t$



6. Suppose that $a_n\to c$ as $n\to\infty$ and that $\left\{\alpha_i\right\}_{i=1}^{\infty}$ is a sequence of positive terms for which $\displaystyle\sum_{i=1}^na_i\to\infty$ as $n\to\infty$.
   1. ($10\%$) Show that

      $\displaystyle\frac{\displaystyle\sum_{i=1}^n\alpha_ia_i}{\displaystyle\sum_{i=1}^n\alpha_i}\to c$ as $n\to\infty$.

      In particular, if $\alpha_i=1$ for all $i$, then

      $\displaystyle\frac1n\sum_{i=1}^na_i\to c$ as $n\to\infty$.

   2. ($5\%$) Show that the converse of the special case in (a) does not always hold by giving a counterexample of a sequence $\left\{a_n\right\}_{n=1}^{\infty}$ that does not converges, yet $\displaystyle\frac1n\sum_{i=1}^na_i$ converges as $n\to\infty$.
訣竅
運用Stolz–Cesàro定理（一種數列版本的羅必達法則）；對於反例可考慮經典的發散數列。
解法
1. 設 $\displaystyle b_n=\sum_{i=1}^n\alpha_i$、$\displaystyle c_n=\sum_{i=1}^n\alpha_ia_i$，按題設可知 $\displaystyle\lim_{n\to\infty}b_n=\infty$。再者，有

   $\displaystyle\lim_{n\to\infty}\frac{c_{n+1}-c_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{\alpha_{n+1}a_{n+1}}{\alpha_{n+1}}=\lim_{n\to\infty}a_{n+1}=c$

   那麼根據Stolz－Cesàro定理可知 $\displaystyle\lim_{n\to\infty}\frac{c_n}{b_n}=c$，此即所欲證。而當 $\alpha_i=1$，那麼容易知道 $b_n=n$，$\displaystyle c_n=\sum_{i=1}^na_i$。
2. 考慮 $a_n=\left(-1\right)^n$，容易直接計算

   $\displaystyle\sum_{i=1}^na_i=\begin{cases}-1,&\text{if}~n~\text{is odd},\\0,&\text{if}~n~\text{is even}.\end{cases}$

   由此可知 $\displaystyle\lim_{n\to\infty}\frac1n\sum_{i=1}^na_i=0$，但數列 $\left\{a_i\right\}_{i=1}^{\infty}$ 發散。