小题合辑

${\Large 1.}$计算极限$$\lim\limits_{n\rightarrow\infty}\frac{[1^p+2^p+\cdots
+(2n-1)^p]^{q+1}}{[1^q+2^q+\cdots+(2n-1)^q]^{p+1}}$$
${\bf 解:}$

\begin{align*}
令&\lim\limits_{n\rightarrow\infty}\frac{[1^p+2^p+\cdots
+(2n-1)^p]^{q+1}}{[1^q+2^q+\cdots+(2n-1)^q]^{p+1}}=e^A\\
则A&=\lim_{n\rightarrow\infty}[(q+1)\ln (1^p+2^p+\cdots+(2n-1)^p)-
(p+1)\ln (1^q+2^q+\cdots+(2n-1)^q)]\\
&=\lim_{n\rightarrow\infty}[(q+1)\ln (n^{p+1}\frac{1^p+2^p+\cdots+(2n-1)^p}{n^{p+1}})-
                                      (p+1)\ln (n^{q+1}\frac{1^q+2^q+\cdots+(2n-1)^q}{n^{q+1}})]\\
&=(q+1)\lim_{n\rightarrow\infty}\ln \frac{1^p+2^p+\cdots+(2n-1)^p}{n^{p+1}}n^{p+1}-
    (p+1)\lim_{n\rightarrow\infty}\ln \frac{1^q+2^q+\cdots+(2n-1)^q}{n^{q+1}}n^{q+1}\\
&=(q+1)\lim_{n\rightarrow\infty}\ln \frac{\sum\limits^n_{i=1}(2\frac{i}{n}-\frac{1}{n})^p}{n}n^{p+1}-
    (p+1)\lim_{n\rightarrow\infty}\ln \frac{\sum\limits^n_{i=1}(2\frac{i}{n}-\frac{1}{n})^q}{n}n^{q+1}\\
&=(q+1)\lim_{n\rightarrow\infty}\ln [\int^1_0(2x)^p{\rm d}x]n^{p+1}-
    (p+1)\lim_{n\rightarrow\infty}\ln [\int^1_0(2x)^q{\rm d}x]n^{q+1}\\
&=(q+1)[\ln [\int^1_0(2x)^p{\rm d}x]+(p+1)\lim_{n\rightarrow\infty}\ln n]-
    (p+1)[\ln [\int^1_0(2x)^q{\rm d}x]+(q+1)\lim_{n\rightarrow\infty}\ln n]\\
&=(q+1)\ln [\int^1_0(2x)^p{\rm d}x]-(p+1)\ln [\int^1_0(2x)^q{\rm d}x]\\
&=(q+1)\ln \frac{2^px^{p+1}}{p+1}|^1_0-(p+1)\ln \frac{2^qx^{q+1}}{q+1}|^1_0\\
&=(p-q)\ln 2+(p+1)\ln (q+1)-(q+1)\ln (p+1)\\
\therefore 原式&=e^{(p-q)\ln 2+(p+1)\ln (q+1)-(q+1)\ln (p+1)}
\end{align*}