Answer:
To find the sum of the series
[tex]S_n = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2},
[/tex]
we can express it as:
[tex]S_n = \sum_{k=1}^{n} \frac{1}{k^2}.[/tex]
This series is known as the partial sum of the Basel problem, which converges to a specific value as n approaches infinity. The exact value of the infinite series is given by:
[tex]\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}.[/tex]
However, since we are interested in the finite sum S_n , we can compute it for specific values of n or use known approximations for larger n .
Step 1: Calculate for Small Values of n
Let's calculate S_n for small values of n :
[tex]- \text{For n = 1 :}[/tex]
[tex]S_1 = 1.[/tex]
[tex]- \text{For} \: n = 2 :[/tex]
[tex]S_2 = 1 + \frac{1}{2^2} = 1 + \frac{1}{4} = \frac{5}{4}.[/tex]
[tex]- \text{For} \: n = 3 :[/tex]
[tex]{S_3 = 1 + \frac{1}{2^2} + \frac{1}{3^2} = 1 + \frac{1}{4} + \frac{1}{9} = \frac{36}{36} + \frac{9}{36} + \frac{4}{36} = \frac{49}{36}.}[/tex]
[tex] \text{For} \: n = 4 :[/tex]
[tex]{S_4 = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} = \frac{49}{36} + \frac{1}{16} = \frac{49}{36} + \frac{9}{144} = \frac{196}{144} + \frac{9}{144} = \frac{205}{144}.}[/tex]
Step 2: General Formula for Large n
For large n , the sum S_n can be approximated using the known result:
[tex]S_n \approx \frac{\pi^2}{6} - \frac{1}{n}.[/tex]
Thus, the final answer for the sum of the series up to n is:
[tex]S_n = \sum_{k=1}^{n} \frac{1}{k^2}.[/tex]
