Solution
= Solution
Write
$$
\frac{j}{j^2+j+1}=\frac1j+O(j^{-2})
$$
uniformly for $j\ge1$. The accumulated error from $j=n+1$ to $2n$ tends to zero, while the corresponding <harmonic sum> tends to $\log2$. Hence $b_n\to\log2$.
= Solution
Write
$$
\frac{j}{j^2+j+1}=\frac1j+O(j^{-2})
$$
uniformly for $j\ge1$. The accumulated error from $j=n+1$ to $2n$ tends to zero, while the corresponding <harmonic sum> tends to $\log2$. Hence $b_n\to\log2$.