Answer: The result is

\(\sum_{n=1}^{N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}=1-\frac{1}{(N+1)(2N+1)}\)

and also

\(\sum_{n=N+1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}\lt \frac{3}{8N^2}.\)

Let

\(H_k:\ \sum_{n=1}^{k} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}=1-\frac{1}{(k+1)(2k+1)}.\)

We prove \(H_k\) by induction.

First, for \(k=1\),

\(\sum_{n=1}^{1} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}=\frac{5}{1\cdot 2\cdot 1\cdot 3}=\frac{5}{6},\)

and

\(1-\frac{1}{(1+1)(2\cdot 1+1)}=1-\frac{1}{6}=\frac{5}{6}.\)

So \(H_1\) is true.

Now assume \(H_k\) is true for some positive integer \(k\). Then

\(\sum_{n=1}^{k+1} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}\)

\(=1-\frac{1}{(k+1)(2k+1)}+\frac{4k+5}{(k+1)(k+2)(2k+1)(2k+3)}.\)

Write the two terms with a common denominator:

\(\frac{1}{(k+1)(2k+1)}-\frac{4k+5}{(k+1)(k+2)(2k+1)(2k+3)}\)

\(=\frac{(k+2)(2k+3)-(4k+5)}{(k+1)(k+2)(2k+1)(2k+3)}\)

\(=\frac{2k^2+3k+1}{(k+1)(k+2)(2k+1)(2k+3)}\)

\(=\frac{(k+1)(2k+1)}{(k+1)(k+2)(2k+1)(2k+3)}\)

\(=\frac{1}{(k+2)(2k+3)}.\)

Hence

\(\sum_{n=1}^{k+1} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}=1-\frac{1}{(k+2)(2k+3)},\)

which is exactly \(H_{k+1}\). Therefore \(H_k\) is true for all positive integers \(k\).

Now apply the result with \(k=2N\) and \(k=N\):

\(\sum_{n=N+1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}\)

\(=\sum_{n=1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}-\sum_{n=1}^{N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}\)

\(=\left(1-\frac{1}{(2N+1)(4N+1)}\right)-\left(1-\frac{1}{(N+1)(2N+1)}\right).\)

So

\(\sum_{n=N+1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}=\frac{1}{(N+1)(2N+1)}-\frac{1}{(2N+1)(4N+1)}.\)

Factorising gives

\(=\frac{3N}{(N+1)(2N+1)(4N+1)}.\)

Since \(N+1\gt N\), \(2N+1\gt 2N\) and \(4N+1\gt 4N\),

\(\frac{3N}{(N+1)(2N+1)(4N+1)}\lt \frac{3N}{N\cdot 2N\cdot 4N}=\frac{3}{8N^2}.\)

Therefore

\(\sum_{n=N+1}^{2N} \frac{4n+1}{n(n+1)(2n-1)(2n+1)}\lt \frac{3}{8N^2}.\)