Question

Suppose a1=12−12,a2=23−13,a3=34−14,a4=45−15,a5=56−16. a) Find an explicit formula for an: . b) Determine whether the sequence is convergent or divergent: . (Enter "convergent" or "divergent" as appropriate.)

Answer 1

The explicit formula for the nth term of the sequence is an = (n+1)n - 1/(n+1). Since the term (n+1)n grows without bound and -1/(n+1) approaches 0 as n approaches infinity, the sequence is divergent.

Step-by-step explanation:

Solution for a) Explicit Formula for an

To find an explicit formula for an, we need to see a pattern in the given sequence. Looking at the given terms:

a1 = 12 - 1/2

a2 = 23 - 1/3

a3 = 34 - 1/4

a4 = 45 - 1/5

a5 = 56 - 1/6

We can observe that each term an has the form (n+1)n - 1/(n+1). Therefore, the explicit formula for the nth term, an, is:

an = (n+1)n - 1/(n+1)

Solution for b) Convergence or Divergence

For the convergence or divergence of the sequence, we consider its behavior as n approaches infinity. The term (n+1)n grows without bound, while the term -1/(n+1) approaches 0. Hence, the sequence grows indefinitely as n becomes larger, which indicates that the sequence is divergent.

Answer 2

I'm going to assume you meant to write fractions (because if
`a_n` are all non-negative integers, the series would clearly diverge), so that

`a_1=\frac12-\frac12`

`a_2=\frac23-\frac13`

`a_3=\frac34-\frac14`

and so on.

a. If the pattern continues as above, we would have the general term

`a_n=\frac n{n+1}-\frac1{n+1}=(n-1)/(n+1)`

b. Note that we can write
`a_n` as

`a_n=(n-1)/(n+1)=(n+1-2)/(n+1)=1-\frac2{n+1}`

The series diverges by comparison to the divergent series

`\displaystyle\sum_(n=1)^\infty\frac1n`