${\large 1.} $ COMPUTE Limit $$\lim\limits_{n\rightarrow\infty}\frac{[1^p+2^p+\cdots
+ (2n-1) ^p]^{q+1}}{[1^q+2^q+\cdots+ (2n-1) ^q]^{p+1}}$$
${\BF Solution:}$
\begin{align*}
Make &\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\\
Then a&=\lim_{n\rightarrow\infty}[(q+1) \ln (1^p+2^p+\cdots+ (2n-1) ^p)-
(q+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 (p+1) \ln (q+1)-(q+1) \ln (p+1) \ \
\therefore original &=e^{(p-q) \ln (p+1) \ln (q+1)-(q+1) \ln (p+1)}
\end{align*}
Small Problem Collection