Question

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

Answer 1

Solution

We use the formula for an:

$a_n=S_n-S_{n-1}$

This is:

$\begin{gathered} =\frac{1}{(n+1)^2}-\frac{1}{n^2} \\ For\text{ 42nd } \\ S_{42}=\frac{1}{(42+1)^2}-\frac{1}{42^2}-\frac{1}{(41+1)^2}+\frac{1}{41^2} \\ S_{42}=\frac{1}{43^2}-\frac{1}{1764}-\frac{1}{42^2}+\frac{1}{1681} \\ S_{42}=\frac{1}{1849}-\frac{1}{1764}-\frac{1}{1764}+\frac{1}{1681} \\ S_{42}=1.93\times10^{-6} \end{gathered}$

Answer:

$1.93\times10^{-6}$