Determine if the series converges or diverges. Summation n=1 to infinity (-1)^n ln[3n+4/3n+3]

Answer 1

Sum from n = 1 to infinity of (-1)^n ln( [3n+4]/[3n+3] )

I have introduced parentheses to make this make more sense. Be very careful with algebraic fractions; use enough parentheses to be clear what is and what is not in the fraction.

This is a tricky problem.

It resembles closely a well-known *telescoping series* which simplifies to ln3 - ln 4 + ln 4 - ln5 + ln 5 ...

Note that by the definition of logarithms (as they are exponents) ln(a/b) = ln a - ln b

So what we really have is

Sum from n = 1 to infinity of (-1)^n ( ln [3n+4] - ln [3n+3] )

Note the (-1)^n gives us alternating -1, +1, -1, +1, . . .

Now let's look at the first several terms of the series n = 1, 2, 3,...

Sum of

1. - ln 7 + ln 6
2. +ln 10 - ln 9
3. - ln 13 - ln 12
4. + ln 16 - ln 15
5. 
   
6. Well, we see that this one does NOT telescope. But remember this approach in case one does.

**************************************

Next consideration. Itis an alternating series. Does it pass the alternating series test?

Three conditions:

(1) limit as n --> infinity of a_n = 0

(2) exactly alternating, +, -, +, -, . . .

(3) For all n, |a_(n+1)| < |a_n| (strictly decreasing in absolute value)

Yes this series passes.

-- We see that as n--> infinity, ([3n+4]/[3n+3]) --> 1

So that ln ([3n+4]/[3n+3]) --> 0

Pass test 1, terms approach zero

-- Signs: ([3n+4]/[3n+3]) > 1 so ln ([3n+4]/[3n+3]) > 0

Therefore the sign is determined by (-1)^n, absolutely alternating.

Passes test 2.

-- Always positive so we don't need to worry about the absolute value.

(3(n+1) + 4) / (3(n+1) + 3) < (3n + 4) / (3n+ 3)

(3n + 7)(3n + 3) < (3n+4)(3n + 6) [OK to multiply, all positive]

9n^2 + 30n + 21 < 9n^2 + 30n + 24

21 < 24

If we reverse the above steps -- read bottom to top -- this proves the n+1'th argument of the ln is always less than the nth.

For u > 0, ln u is an increasing function; that is, if b > a, ln b > ln a and if b < a, ln b < ln a.

Since the fraction is (as shown above) a strictly decreasing function, then so is ln ([3n+4]/[3n+3])

Passes test 3.

***By the Alternating Series Test, this series **converges.**

If it passes as an alternating series but bnot in absolute value, we say it converges *conditionally*

If the absolute value converges, we say it converges *absolutely*. I am not sure if this converges absolutely or not, but we e=were not asked that, so let it go.