Question

Determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in Example 8). [infinity] 6 n2 − 1 n = 3 convergent divergent. If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.)

Answer 1

Answer:

Step-by-step explanation:

given

$\sum\limits^\infty_{n=3}\frac{6}{n^2-1}=\sum\limits^\infty_{n=3}\frac{6}{(n-1)(n+1)}\\\\=\sum\limits^\infty_{n=3}\frac{6}{2}(\frac{1}{n-1}-\frac{1}{n+1})\\\\=\frac{6}{2}\sum\limits^\infty_{n=3}(\frac{1}{n-1}-\frac{1}{n+1})\\\\=3[(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})+...]\\\\=3[\frac{1}{2}+\frac{1}{3}]=3[\frac{3+2}{6}]=3[\frac{5}{6}]\\\\=\frac{15}{6}=\frac{5}{3}$