1. **Planteamiento del problema:** Calcular las primeras 10 sumas parciales de la serie $$\sum_{n=1}^{\infty} \ln \left( \frac{n^2 + 1}{2n^2 + 1} \right)$$. 2. **Fórmula y reglas importantes:** La suma parcial $$S_k$$ es la suma de los primeros $$k$$ términos: $$ S_k = \sum_{n=1}^k \ln \left( \frac{n^2 + 1}{2n^2 + 1} \right) $$ Recordemos que $$\ln(a) + \ln(b) = \ln(ab)$$, por lo que podemos escribir: $$ S_k = \ln \left( \prod_{n=1}^k \frac{n^2 + 1}{2n^2 + 1} \right) $$ 3. **Cálculo de las primeras 10 sumas parciales:** Calculamos cada término y luego la suma parcial: Para $$n=1$$: $$\ln \left( \frac{1^2 + 1}{2(1)^2 + 1} \right) = \ln \left( \frac{2}{3} \right)$$ Para $$n=2$$: $$\ln \left( \frac{4 + 1}{8 + 1} \right) = \ln \left( \frac{5}{9} \right)$$ Para $$n=3$$: $$\ln \left( \frac{9 + 1}{18 + 1} \right) = \ln \left( \frac{10}{19} \right)$$ Para $$n=4$$: $$\ln \left( \frac{16 + 1}{32 + 1} \right) = \ln \left( \frac{17}{33} \right)$$ Para $$n=5$$: $$\ln \left( \frac{25 + 1}{50 + 1} \right) = \ln \left( \frac{26}{51} \right)$$ Para $$n=6$$: $$\ln \left( \frac{36 + 1}{72 + 1} \right) = \ln \left( \frac{37}{73} \right)$$ Para $$n=7$$: $$\ln \left( \frac{49 + 1}{98 + 1} \right) = \ln \left( \frac{50}{99} \right)$$ Para $$n=8$$: $$\ln \left( \frac{64 + 1}{128 + 1} \right) = \ln \left( \frac{65}{129} \right)$$ Para $$n=9$$: $$\ln \left( \frac{81 + 1}{162 + 1} \right) = \ln \left( \frac{82}{163} \right)$$ Para $$n=10$$: $$\ln \left( \frac{100 + 1}{200 + 1} \right) = \ln \left( \frac{101}{201} \right)$$ Sumamos los términos para obtener las sumas parciales: $$ S_1 = \ln \left( \frac{2}{3} \right) \approx -0.4055 $$ $$ S_2 = \ln \left( \frac{2}{3} \times \frac{5}{9} \right) = \ln \left( \frac{10}{27} \right) \approx -0.9933 $$ $$ S_3 = \ln \left( \frac{10}{27} \times \frac{10}{19} \right) = \ln \left( \frac{100}{513} \right) \approx -1.6350 $$ $$ S_4 = \ln \left( \frac{100}{513} \times \frac{17}{33} \right) = \ln \left( \frac{1700}{16929} \right) \approx -2.2730 $$ $$ S_5 = \ln \left( \frac{1700}{16929} \times \frac{26}{51} \right) = \ln \left( \frac{44200}{863379} \right) \approx -3.0700 $$ $$ S_6 = \ln \left( \frac{44200}{863379} \times \frac{37}{73} \right) = \ln \left( \frac{1635400}{63026867} \right) \approx -4.1430 $$ $$ S_7 = \ln \left( \frac{1635400}{63026867} \times \frac{50}{99} \right) = \ln \left( \frac{81770000}{6239655783} \right) \approx -4.9990 $$ $$ S_8 = \ln \left( \frac{81770000}{6239655783} \times \frac{65}{129} \right) = \ln \left( \frac{5315050000}{804312597507} \right) \approx -5.9990 $$ $$ S_9 = \ln \left( \frac{5315050000}{804312597507} \times \frac{82}{163} \right) = \ln \left( \frac{435034100000}{131901954540541} \right) \approx -6.9990 $$ $$ S_{10} = \ln \left( \frac{435034100000}{131901954540541} \times \frac{101}{201} \right) = \ln \left( \frac{43938444100000}{26512292353035541} \right) \approx -7.9990 $$ 4. **Interpretación:** Las sumas parciales decrecen y se aproximan a un valor negativo grande. **Respuesta final:** Las primeras 10 sumas parciales son aproximadamente: $$ S_1 \approx -0.4055, S_2 \approx -0.9933, S_3 \approx -1.6350, S_4 \approx -2.2730, S_5 \approx -3.0700, S_6 \approx -4.1430, S_7 \approx -4.9990, S_8 \approx -5.9990, S_9 \approx -6.9990, S_{10} \approx -7.9990 $$