Series of 6/(3n+1)(3n+4)

[SamuelFreidson]

The sequence is [imath]\frac{6}{4 * 7} + \frac{6}{7 * 10} + \frac{6}{10 * 13} + ...[/imath], so I assume that it's something to do with canceling the like terms. I'm not sure how to do that with multiplication though; only addition. We're supposed to do it by finding some [imath]s_n[/imath] to represent the sum at n and then find [imath]lim_{n\rightarrow\infty}s_n[/imath].

 

[khansaheb]

View attachment 37579
The sequence is [imath]\frac{6}{4 * 7} + \frac{6}{7 * 10} + \frac{6}{10 * 13} + ...[/imath], so I assume that it's something to do with canceling the like terms. I'm not sure how to do that with multiplication though; only addition. We're supposed to do it by finding some [imath]s_n[/imath] to represent the sum at n and then find [imath]lim_{n\rightarrow\infty}s_n[/imath].

Start with:

6 / [(3n+1) * (3n + 4)] = 2* [1/(3n+1) - 1/(3n+4)] ............... now do the summation for two of those sequences.