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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.)

Respuesta :

fortuneonyemuwa

Answer:

Step-by-step explanation:

given

[tex]\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}[/tex]

jayilych4real

We can determine that the series is convergent and the sum is 15/6

What is a Series?

This is a group of numbers that are added together to find infinite qualities one after the other to a given starting quantity.

Hence, we can see that

Σ to infinity, from n=3

6/2 (1/n-1) - (1/n+1)

When we expand,

3[(1/2 + 1/3)] = 3[3+2/6]

= 3[5/6]

=15/6

Read more about convergent series here:

https://brainly.com/question/10813422

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