1. **Problem:** Simplify the finite sum $$\sum_{n=1}^N \frac{1}{(n+1)(n+2)}$$ and show that $$\sum_{n=1}^\infty \frac{1}{(n+1)(n+2)} = \frac{1}{2}.$$\n\n2. **Formula and approach:** Use partial fraction decomposition to simplify the general term. Recall that for telescoping sums, terms cancel out in sequence.\n\n3. **Partial fraction decomposition:**\n$$\frac{1}{(n+1)(n+2)} = \frac{A}{n+1} + \frac{B}{n+2}$$\nMultiply both sides by $(n+1)(n+2)$:\n$$1 = A(n+2) + B(n+1) = (A+B)n + (2A + B)$$\nEquate coefficients:\n$$A + B = 0 \quad \Rightarrow \quad B = -A$$\n$$2A + B = 1 \quad \Rightarrow \quad 2A - A = 1 \Rightarrow A = 1$$\nThus, $$A=1, B=-1$$ and\n$$\frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2}.$$\n\n4. **Rewrite the sum:**\n$$\sum_{n=1}^N \frac{1}{(n+1)(n+2)} = \sum_{n=1}^N \left( \frac{1}{n+1} - \frac{1}{n+2} \right)$$\n\n5. **Telescoping effect:**\nWrite out first few terms:\n$$\left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \cdots + \left( \frac{1}{N+1} - \frac{1}{N+2} \right)$$\nMost terms cancel, leaving:\n$$= \frac{1}{2} - \frac{1}{N+2}.$$\n\n6. **Limit as $N \to \infty$:**\n$$\lim_{N \to \infty} \sum_{n=1}^N \frac{1}{(n+1)(n+2)} = \lim_{N \to \infty} \left( \frac{1}{2} - \frac{1}{N+2} \right) = \frac{1}{2}.$$\n\n---\n\n7. **Problem:** Test convergence of the series:\ni. $$\sum_{n=1}^\infty \frac{4^n}{n 5^n}$$\nii. $$\sum_{n=1}^\infty \frac{n^4}{n^5 + n}$$\n\n8. **Test i:** Simplify the term:\n$$\frac{4^n}{n 5^n} = \frac{1}{n} \left( \frac{4}{5} \right)^n.$$\nSince $\frac{4}{5} < 1$, this is a positive term series with terms decreasing exponentially.\nUse the root or ratio test:\n$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{1}{n+1} \left( \frac{4}{5} \right)^{n+1} \cdot \frac{n}{\left( \frac{4}{5} \right)^n} = \frac{4}{5} < 1.$$\nTherefore, the series converges absolutely.\n\n9. **Test ii:** Simplify the term:\n$$\frac{n^4}{n^5 + n} = \frac{n^4}{n(n^4 + 1)} = \frac{n^4}{n (n^4 + 1)} = \frac{n^3}{n^4 + 1}.$$\nFor large $n$, $$\frac{n^3}{n^4 + 1} \approx \frac{n^3}{n^4} = \frac{1}{n}.$$\nSince $$\sum \frac{1}{n}$$ diverges (harmonic series), compare test shows this series diverges.\n\n**Final answers:**\n- $$\sum_{n=1}^\infty \frac{1}{(n+1)(n+2)} = \frac{1}{2}.$$\n- Series i converges absolutely.\n- Series ii diverges.