Approximation of Sum of Partial Harmonic series with only odd terms with elementary methods 
Let $a=\frac{1^2}{1}+\frac{2^2}{3}+\frac{3^2}{5}…+\frac{1001^2}{2001}$ and $b=\frac{1^2}{3}+\frac{2^2}{5}+\frac{3^2}{7}…+\frac{1001^2}{2003}$ Find the closest integer to $a-b$

Using the identity $a^2-b^2=(a+b)(a-b)$, it was trivial to see that the above expression was equal to $$1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}…+\frac{1}{2001}-\frac{1001^2}{2003}$$
I know that the approximation for the summation of a Harmonic series till $n$ terms is $\ln(n)<S_{n}<\ln(n)+1$ but I couldn't figure out how to use that to arrive at an approximation of the above expression.
EDIT: I just made a really dumb mistake which somehow I didn't notice before posting. The expression I have written is wrong, @cosmo5 has corrected it in his answer. Nevertheless, I would still like to know how to approximate a Partial Harmonic Series, so I'm leaving my post as is
PS- Questions on Harmonic series always have answers that use advanced math, and I don't know any of that so I would request you to refrain from using it. I just want an approximation with an error margin of $≤1$ which the elementary methods should be good enough for
 A: Without Harmonic series approximation,
$a-b=$
$$1^2 + \frac{2^2-1^2}{3} + \frac{3^2-2^2}{5} + \ldots + \frac{1001^2-1000^2}{2001} - \frac{1001^2}{2003}$$
$$=1 + (2-1) + (3-2) + \ldots + (1001-1000) -\frac{1001^2}{2003}$$
$$=1001 - \frac{1001^2}{2003}$$
$$=1001\cdot\frac{1002}{2003}$$
$$\approx \boxed{501}$$
A: $$\sum _{k=1}^{n} \left(\frac{k^2}{2 k-1}-\frac{k^2}{2 k+1}\right)=\sum _{k=1}^n \frac{2 k^2}{4 k^2-1}=\frac{n^2+n}{2 n+1}$$
for $n=1001$ the difference is $500.75$
The answer is $501$
For the approximation of the partial harmonic numbers sum, I got
$$\sum _{k=1}^{n} \frac{1}{2 k-1}\approx\frac{1}{2} (\log (n)+\gamma +2 \log (2))$$
Where $\gamma\approx 0.5772156649$ is the Euler-Mascheroni constant.
It works great, because for $n=10000$ I got an error less than $10^9$.
Edit
Adding a term to the expansion at $\infty$ of the actual sum we have much more precision
$$\sum _{k=1}^{n} \frac{1}{2 k-1}\approx \frac{1}{48} \left(\frac{1}{n^2}+24 (\log (4 n)+\gamma )\right)$$
For $n=100$ I got an error less than $10^{10}$.
