我遇到了一个问题，不同的计算机编译文件的方式不同。在我的台式计算机上，一切都像正常一样编译，但在我的新笔记本电脑上，底部边距被切掉了。我在两台计算机上都使用 TeXstudio 和 MiKTeX，并且我仔细检查了两台计算机是否都设置为在 MiKTeX 中使用标准信纸大小的纸张，所以这似乎不是问题。
我真的很困惑，我以前从未遇到过这个问题。感谢您提供的任何帮助！
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\lhead{MTH 2410}
\chead{Summary of Tests for Series}
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\begin{table}[h!]
\centering
\begin{tabular}{C{2.9cm} C{2.6cm} C{3.46cm} C{3.2cm} L{3.5cm}}
\toprule
\textbf{Series or Test} & \textbf{Form of Series} & \textbf{Condition(s) for Convergence} & \textbf{Condition(s) for Divergence} & \textbf{Comments} \\
\midrule
Geometric Series & $\sum_{k=N}^{\infty} ar^k$, $a\ne 0$ \newline[1mm] $N=0,1,2,\dots$ & $|r|<1$ & $|r|\geq 1$ & If $|r|<1$, then the \newline[1mm] sum is $S=\frac{ar^N}{1-r}$. \\
\midrule
Telescoping Series & $\sum_{k=1}^{\infty}\big(b_k - b_{k+1}\big)$ & $\lim_{n\to \infty} S_n = L$, where \newline[1mm] $-\infty<L<\infty$ & $\lim_{n\to \infty} S_n$ DNE & Trick: Write out $S_n$, cancel terms, then evaluate $\lim_{n\to\infty} S_n$. \\
\midrule
Divergence & $\sum_{k=1}^{\infty} a_k$ & This test \underline{\textbf{CANNOT}} be used to show convergence! & $\lim_{k\to \infty} a_k \ne 0$ & Test is \textbf{inconclusive} if $\lim_{k\to\infty} a_k=0$. \\
\midrule
Integral ($f$ is continuous, positive, and decreasing) & $\sum_{k=1}^{\infty} a_k$, where \newline[1mm] $0<a_k=f(k)$ & $\int_1^{\infty}\! f(x)\, dx$ converges \newline[2mm] The series does \underline{\textbf{NOT}} converge to this value. & $\int_1^{\infty}\! f(x)\, dx$ diverges & Remainder satisfies \newline[1mm] $|R_n| \leq \int_n^{\infty}\! f(x)\, dx$ \\
\midrule
$p$-Series & $\sum_{k=1}^{\infty} \frac{1}{k^p}$ & $p>1$ & $p\leq 1$ & Useful for comparison tests. \\
\midrule
Ratio & $\sum_{k=1}^{\infty} a_k$ & $\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|<1$ & $\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|>1$ & Test is \textbf{inconclusive} \newline[1mm] if $\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=1$. \\
\midrule
Root & $\sum_{k=1}^{\infty} a_k$ & $\lim_{k\to\infty}\sqrt[k]{|a_k|} < 1$ & $\lim_{k\to\infty}\sqrt[k]{|a_k|} > 1$ & Test is \textbf{inconclusive} \newline[1mm] if $\lim_{k\to\infty}\sqrt[k]{|a_k|} = 1$. \\
\midrule
Direct Comparison ($a_k,b_k>0$) & $\sum_{k=1}^{\infty} a_k$ & $0<a_k\leq b_k$ and \newline[1mm] $\textstyle\Sigma b_k$ converges. & $0<b_k\leq a_k$ and \newline[1mm] $\textstyle\Sigma b_k$ diverges. & $\textstyle \Sigma a_k$ is given; you supply $\textstyle \Sigma b_k$. \newline[1mm] The ``work'' is showing the inequality. \\
\midrule
Limit Comparison ($a_k,b_k>0$) & $\sum_{k=1}^{\infty} a_k$ & $\lim_{k\to\infty}\frac{a_k}{b_k} = L$, \newline[1mm] $0\leq L<\infty$, and \newline[1mm] $\textstyle\Sigma b_k$ converges. & $\lim_{k\to\infty}\frac{a_k}{b_k} = L$, \newline[1mm] $0<L\leq\infty$, and \newline[1mm] $\textstyle\Sigma b_k$ diverges. & $\textstyle \Sigma a_k$ is given; you supply $\textstyle \Sigma b_k$. \\
\midrule
Alternating \newline Series & $\sum_{k=1}^{\infty} (-1)^{k}a_k$, \newline[1mm] where $a_k> 0$ & $\lim_{k\to\infty} a_k = 0$ and \newline[1mm] $0<a_{k+1}\leq a_k$ & $\lim_{k\to\infty} a_k \ne 0$ & Remainder satisfies \newline[1mm] $|R_n|\leq a_{n+1}$ \\
\midrule
Absolute Convergence & $\sum_{k=1}^{\infty} a_k$ & $\sum_{k=1}^{\infty} |a_k|$ converges & n/a & If $\textstyle \Sigma |a_k|$ diverges, $\textstyle \Sigma a_k$ may be conditionally convergent. \\
\bottomrule
\end{tabular}
\end{table}
}
\underline{\textbf{Properties of Convergent Series}}\\[2mm]
Suppose that $\textstyle \sum a_k$ and $\textstyle \sum b_k$ converge to $A$ and $B$, respectively, and let $c$ be any real number.
\begin{enumerate}[noitemsep]
\item The series $\textstyle \sum ca_k$ converges and $\textstyle \sum ca_k = c \sum a_k = cA$.
\item The series $\textstyle \sum (a_k \pm b_k)$ converges and $\textstyle \sum (a_k \pm b_k) = \sum a_k \pm \sum b_k = A \pm B$.
\item \textit{Whether} a series converges does not depend on a finite number of terms added to or removed from the series. However, the \textit{value} of a convergent series does change if nonzero terms are added or deleted.
\end{enumerate}
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