$$ \left\{\begin{array}{l}a=\frac{1}{1 imes 2}+\frac{1}{3 imes 4}+...+\frac{1}{1403 imes 1404} \b=\frac{1}{703 imes 1404}+\frac{1}{704 imes 1403}+...+\frac{1}{1404 imes 703} \\frac{a}{b}=\end{array} ight. $$

Question

$$ \left\{\begin{array}{l}a=\frac{1}{1	imes 2}+\frac{1}{3	imes 4}+...+\frac{1}{1403	imes 1404} \b=\frac{1}{703	imes 1404}+\frac{1}{704	imes 1403}+...+\frac{1}{1404	imes 703} \\frac{a}{b}=\end{array}ight. $$

UpStudy AI · Accepted Answer

We first analyze the sum for $$a$$. Notice that
$$
a=\frac{1}{1\times2}+\frac{1}{3\times4}+\cdots+\frac{1}{1403\times1404}
$$
can be written as
$$
a=\sum_{k=1}^{702}\frac{1}{(2k-1)(2k)}.
$$
Observe that
$$
\frac{1}{(2k-1)(2k)}=\frac{1}{2k-1}-\frac{1}{2k},
$$
since
$$
\frac{1}{2k-1}-\frac{1}{2k}=\frac{2k-(2k-1)}{(2k-1)(2k)}=\frac{1}{(2k-1)(2k)}.
$$
Thus, the sum becomes
$$
a=\sum_{k=1}^{702}\left(\frac{1}{2k-1}-\frac{1}{2k}\right).
$$
We can split the sum as
$$
a=\left(\frac{1}{1}+\frac{1}{3}+\cdots+\frac{1}{1403}\right)
-\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{1404}\right).
$$
Notice that the sum of the even reciprocals is
$$
\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{1404}=\frac{1}{2}\sum_{k=1}^{702}\frac{1}{k}.
$$
The sum of the odd reciprocals can be expressed using the full harmonic sum. In fact, the full harmonic sum up to 1404 is
$$
H_{1404}=\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{1404},
$$
so the odd terms sum is
$$
\frac{1}{1}+\frac{1}{3}+\cdots+\frac{1}{1403}=H_{1404}-\frac{1}{2}\sum_{k=1}^{702}\frac{1}{k}.
$$
Therefore,
$$
a=\left(H_{1404}-\frac{1}{2}\sum_{k=1}^{702}\frac{1}{k}\right)-\frac{1}{2}\sum_{k=1}^{702}\frac{1}{k}=H_{1404}-\sum_{k=1}^{702}\frac{1}{k}.
$$
This difference is exactly
$$
H_{1404}-H_{702}=\sum_{k=703}^{1404}\frac{1}{k}.
$$
Hence,
$$
a=\sum_{k=703}^{1404}\frac{1}{k}.
$$

Next, consider the sum for $$b$$:
$$
b=\frac{1}{703\times1404}+\frac{1}{704\times1403}+\cdots+\frac{1}{1404\times703}.
$$
We note that each term is of the form 
$$
\frac{1}{k(2107-k)},
$$
with $$k$$ running from 703 to 1404, since $$703+1404=2107$$, $$704+1403=2107$$, and so on. Each term can be decomposed as follows:
$$
\frac{1}{k(2107-k)}=\frac{1}{2107}\left(\frac{1}{k}+\frac{1}{2107-k}\right).
$$
Thus, the sum becomes
$$
b=\sum_{k=703}^{1404}\frac{1}{2107}\left(\frac{1}{k}+\frac{1}{2107-k}\right)
=\frac{1}{2107}\sum_{k=703}^{1404}\left(\frac{1}{k}+\frac{1}{2107-k}\right).
$$
Since as $$k$$ runs from 703 to 1404, the expression $$2107-k$$ also runs from 703 to 1404, we have
$$
\sum_{k=703}^{1404}\left(\frac{1}{k}+\frac{1}{2107-k}\right)=2\sum_{k=703}^{1404}\frac{1}{k}.
$$
Therefore,
$$
b=\frac{2}{2107}\sum_{k=703}^{1404}\frac{1}{k}.
$$
Recalling that
$$
a=\sum_{k=703}^{1404}\frac{1}{k},
$$
we obtain
$$
b=\frac{2a}{2107}.
$$

Finally, we compute $$\frac{a}{b}$$:
$$
\frac{a}{b}=\frac{a}{\frac{2a}{2107}}=\frac{2107}{2}.
$$

Thus, the final result is
$$
\frac{a}{b}=\frac{2107}{2}.
$$
$$ \frac{a}{b} = \frac{2107}{2} $$

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