# Thread: Sequence and bounds (2)

1. ## Sequence and bounds (2)

can some one prove to me why

$(1.2)^{-n}$ is a decreasing sequence?

I know it's obvious but in my textbook they prove it by finding the derivative

which should be $\frac{-(1.2)^{-n}}{ln1.2}$ which is increasing though right?

2. Surely the derivative is just $(-n)(1.2)^{-n-1}$? Which is negative for $n>0$, hence decreasing? Unless I'm misreading the question as I've no clue how logs appeared...

But your derivative is also always negative; either way it's decreasing.

3. I don't think u could do that...or I'm pretty sure...because the variable is a power not a base

4. Oh right that was stupid of me, sorry! I redid it and got the same as you. log(1.2) is positive and $(1.2)^n$ is also positive, so you definitely have a negative answer which shows decreasing. I don't see the problem.

5. no but $(1.2)^{-n}$ gets thrown to the denominator right? and theres already a negative sign in front of it
doesn't that make it increase to 0?

6. Yes as n tends to infinity the derivative tends to 0, but it is never actually 0, it just gets closer and closer. Any value of n you choose will give a negative answer.

7. Originally Posted by akhayoon
can some one prove to me why

$(1.2)^{-n}$ is a decreasing sequence?

I know it's obvious but in my textbook they prove it by finding the derivative

which should be $\frac{-(1.2)^{-n}}{ln1.2}$ which is increasing though right?
Your sequence is obtained by sampling $f(x)=1.2^{-x}$ this function is decreasing because its derivative is negative. Because the function is decreasing the sequence obtained by sampling it at $1, 2, ..$ is also decreasing.

(you cannot differentiate a sequence because the index is a discrete variable and you need a continuous variable to differentiate)

RonL