这是一份leeds利兹大学MATH223001/MATH223101作业代写的成功案例
We apply the lower bound in Lemma 2.5.1 to the logarithms. For a typical term, we get
$$
\ln \left(\frac{m+t-k}{m-k}\right) \geq \frac{\frac{m+t-k}{m-k}-1}{\frac{m+t-k}{m-k}}=\frac{t}{m+t-k}
$$
and so
$$
\begin{aligned}
\ln \left(\frac{m+t}{m}\right) &+\ln \left(\frac{m+t-1}{m-1}\right)+\cdots+\ln \left(\frac{m+1}{m-t+1}\right) \
& \geq \frac{t}{m+t}+\frac{t}{m+t-1}+\cdots+\frac{t}{m+1} .
\end{aligned}
$$
We replace each denominator by the largest one to decrease the sum:
$$
\frac{t}{m+t}+\frac{t}{m+t-1}+\cdots+\frac{t}{m+1} \geq \frac{t^{2}}{m+t} .
$$
MATH223001/MATH223101 COURSE NOTES ：
For $n=1$ and $n=2$, if we require that $L_{n}$ be of the given form, then we get
$$
L_{1}=1=a+b, \quad L_{2}=3=a \frac{1+\sqrt{5}}{2}+b \frac{1-\sqrt{5}}{2}
$$
Solving for $a$ and $b$, we get
$$
a=\frac{1+\sqrt{5}}{2}, \quad b=\frac{1-\sqrt{5}}{2}
$$
Then
$$
L_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}+\left(\frac{1-\sqrt{5}}{2}\right)^{n}
$$