I'm going to assume you meant to write fractions (because if [tex]a_n[/tex] are all non-negative integers, the series would clearly diverge), so that
[tex]a_1=\dfrac12-\dfrac12[/tex]
[tex]a_2=\dfrac23-\dfrac13[/tex]
[tex]a_3=\dfrac34-\dfrac14[/tex]
and so on.
a. If the pattern continues as above, we would have the general term
[tex]a_n=\dfrac n{n+1}-\dfrac1{n+1}=\dfrac{n-1}{n+1}[/tex]
b. Note that we can write [tex]a_n[/tex] as
[tex]a_n=\dfrac{n-1}{n+1}=\dfrac{n+1-2}{n+1}=1-\dfrac2{n+1}[/tex]
The series diverges by comparison to the divergent series
[tex]\displaystyle\sum_{n=1}^\infty\frac1n[/tex]
The pattern follows a sequence other than arithmetic and geometric sequence.
(a) The explicit formula
The given parameters are:
[tex]\mathbf{a_1 = \frac 12 - \frac 12}[/tex]
[tex]\mathbf{a_2 = \frac 23 - \frac 13}[/tex]
[tex]\mathbf{a_3 = \frac 34 - \frac 14}[/tex]
Express 4 as 3 + 1
[tex]\mathbf{a_3 = \frac 3{3+1} - \frac 1{3+1}}[/tex]
Substitute n for 3
[tex]\mathbf{a_n = \frac n{n+1} - \frac 1{n+1}}[/tex]
Hence, the explicit formula is:
[tex]\mathbf{a_n = \frac n{n+1} - \frac 1{n+1}}[/tex]
(b) The sequence type
In (a), we have:
[tex]\mathbf{a_n = \frac n{n+1} - \frac 1{n+1}}[/tex]
Take LCM
[tex]\mathbf{a_n = \frac {n -1}{n+1} }[/tex]
Express -1 as 1 - 2
[tex]\mathbf{a_n = \frac {n +1 - 2}{n+1} }[/tex]
Split
[tex]\mathbf{a_n = \frac {n +1}{n+1} - \frac{2}{n+1} }[/tex]
[tex]\mathbf{a_n =1 - \frac{2}{n+1} }[/tex]
[tex]\mathbf{- \frac{2}{n+1} }[/tex] implies that, the sequence diverges.
Hence, the sequence is a divergent sequence
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