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

SamuelFreidson

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The sequence is 647+6710+61013+...\frac{6}{4 * 7} + \frac{6}{7 * 10} + \frac{6}{10 * 13} + ..., 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 sns_n to represent the sum at n and then find limnsnlim_{n\rightarrow\infty}s_n.
 
Start with:

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