Question

Find the 42nd partial sumS42of the series∞∑n=1 (1/(n+1)^2-1/n^2).

Answer 1

We use the formula for an:

`a_n=S_n-S_(n-1)`

This is:

`\begin{gathered} =(1)/((n+1)^2)-(1)/(n^2) \\ For\text{ 42nd } \\ S_(42)=(1)/((42+1)^2)-(1)/(42^2)-(1)/((41+1)^2)+(1)/(41^2) \\ S_(42)=(1)/(43^2)-(1)/(1764)-(1)/(42^2)+(1)/(1681) \\ S_(42)=(1)/(1849)-(1)/(1764)-(1)/(1764)+(1)/(1681) \\ S_(42)=1.93*10^(-6) \end{gathered}`

Answer:

`1.93*10^(-6)`