Question

Determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in this example). [infinity] 6 n(n + 3) n = 1

a) convergent

b) divergent

If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.)

Answer 1

The series
`\(\sum (6)/(n(n + 3))\)` is convergent as a telescoping sum, and its sum is
`\((11)/(3)\)` or approximately 3.67.

Step-by-step explanation:

The question asks to determine whether the series
`\(\sum (6)/(n(n + 3))\)` from n = 1 to infinity converges or diverges. To solve this, we will express the general term as a difference of two fractions that will telescope when we add the terms of the series.

To start, notice that n(n + 3) can be written as (n + 3) - n. This allows us to decompose the general term an =
`(6)/(n(n + 3))` using partial fraction decomposition:

`\[(6)/(n(n + 3)) = (A)/(n) + (B)/(n + 3)\]`

By finding constants A and B (which turn out to be A = 2 and B = -2), the series becomes:

`\[(2)/(n) - (2)/(n + 3)\]`

As we sum an from n = 1 to infinity, the terms will cancel out in a telescoping fashion. The sum becomes:

`\[S = ((2)/(1) - (2)/(4)) + ((2)/(2) - (2)/(5)) + ((2)/(3) - (2)/(6)) + \ldots\]`

Most terms will cancel, and what we're left with are the first three terms of the numerator: 2 + 1 +
`(2)/(3),` which equals
`(11)/(3)` or approximately 3.67. Thus, the series is convergent and its sum is
`(11)/(3).`

Answer 2

I assume the series is supposed to be

`\displaystyle\sum_(n=1)^\infty \frac6{n(n+3)}`

The summand can be expanded into partial fractions:

`\frac6{n(n+3)}=\frac an+\frac b{n+3}`

`\implies 6=a(n+3)+bn=(a+b)n+3a`

`\implies\begin{cases}a+b=0\\3a=6\end{cases}\implies a=2,b=-2`

Then the sum is equivalent to

`\displaystyle2\sum_(n=1)^\infty\left(\frac1n-\frac1{n+3}\right)`

Consider the N-th partial sum of the series:

2[(1 - 1/4) + (1/2 - 1/5) + (1/3 - 1/6) + (1/4 - 1/7) + (1/5 - 1/8) + …

+ (1/(N - 3) - 1/N) + (1/(N - 2) - 1/(N + 1)) + (1/(N - 1) - 1/(N + 2)) + (1/N - 1/(N + 3))]

Each of the negative terms (in bold) will have a positive counterpart (underlined) that cancel, so the N-th partial sum would be

`\displaystyle\sum_(n=1)^N\frac6{n(n+3)} = 2\left(1+\frac12+\frac13-\frac1{N+1}-\frac1{N+2}-\frac1{N+3}\right)`

As N approaches infinity, the last three terms in the sum converge to 0, so the original sum converges to

`\displaystyle\sum_(n=1)^\infty \frac6{n(n+3)} = 2\left(1+\frac12+\frac13\right) = \boxed{\frac{11}3}`