Answer:

The term can be written as a difference of two consecutive-looking terms:

\(a_n=\frac{n+5}{\sqrt{n^{2}-n+1}}-\frac{n+6}{\sqrt{n^{2}+n+1}}.\)

Notice that

\(n^{2}+n+1=(n+1)^{2}-(n+1)+1.\)

So the second part of \(a_n\) cancels with the first part of \(a_{n+1}\). Therefore the sum telescopes:

\(S_N=\frac{6}{\sqrt{1}}-\frac{N+6}{\sqrt{N^{2}+N+1}}=6-\frac{N+6}{\sqrt{N^{2}+N+1}}.\)

For \(N=30\),

\(S_{30}=6-\frac{36}{\sqrt{30^{2}+30+1}}=6-\frac{36}{\sqrt{931}}=4.820\quad\text{to 3 d.p.}\)

For \(S_N\gt 4.9\),

\(6-\frac{N+6}{\sqrt{N^{2}+N+1}}\gt 4.9,\)

so

\(\frac{N+6}{\sqrt{N^{2}+N+1}}\lt 1.1.\)

Squaring both sides gives

\((N+6)^{2}\lt 1.21(N^{2}+N+1).\)

This simplifies to

\(21N^{2}-1079N-3479\gt 0.\)

Checking the first integer above the positive root gives \(N=55\), and this is the least value.