Determine whether the sequence converges or diverges. If it converges, find the limit. (If an answer does not exist, enter DNE.) 1 2 , 1 3 , 1 3 , 1 4 , 1 4 , 1 5 , 1 5 , 1 6 , ...

if the sequence is:

12, 13, 13, 14, 14 etc, and each term keeps growing up, the sequence obviusly diverges.

Now, if the sequence is

1/2, 1/3, 1/3, 1/4, 1/4, 1/5 , 1/5

so the terms after the first one repeat, we could group the terms with the same denominator and get:

1/2, 2/3, 2/4, 2/5..... etc.

So the terms after the first one are aₙ = 2/n.

Now, a criteria to see if a sequence converges if seing if:

and here we have;

that obviusly tends to zero, so we can conclude that this sequence converges.

then the limit is:

There exist a n' such that for any n > n' then IL -aₙI < ε

where L is the limit

I2/n - 0I = I2/nI < ε

then this is true if n > 2/ε = n'

step-by-step explanation:

i believe its "c" because all of the other options include a but it could be d because there is a way of financing with cash at certain places but its uncommon.

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.