76. ∑(i = 1 to ∞) ((1 / (3i + 1)) - (1 / (3i + 4)))

Explanation:

Step1: Expand the series

Let's write out the first - few terms of the series. When (i = 1), the term is (\frac{1}{3\times1 + 1}-\frac{1}{3\times1+4}=\frac{1}{4}-\frac{1}{7}); when (i = 2), the term is (\frac{1}{3\times2 + 1}-\frac{1}{3\times2+4}=\frac{1}{7}-\frac{1}{10}); when (i = 3), the term is (\frac{1}{3\times3 + 1}-\frac{1}{3\times3+4}=\frac{1}{10}-\frac{1}{13}); and in general, for the (n) - th partial sum (S_n=\sum_{i = 1}^{n}\left(\frac{1}{3i + 1}-\frac{1}{3i+4}\right)).

Step2: Observe the telescoping property

We can see that most of the terms will cancel out. (S_n=\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{13}\right)+\cdots+\left(\frac{1}{3n + 1}-\frac{1}{3n+4}\right)). After cancellation, we are left with (S_n=\frac{1}{4}-\frac{1}{3n + 4}).

Step3: Find the sum of the infinite series

We know that (\sum_{i = 1}^{\infty}\left(\frac{1}{3i + 1}-\frac{1}{3i+4}\right)=\lim_{n\rightarrow\infty}S_n). (\lim_{n\rightarrow\infty}S_n=\lim_{n\rightarrow\infty}\left(\frac{1}{4}-\frac{1}{3n + 4}\right)). As (n\rightarrow\infty), (\lim_{n\rightarrow\infty}\frac{1}{3n + 4}=0). So (\lim_{n\rightarrow\infty}S_n=\frac{1}{4}).

Answer:

(\frac{1}{4})