How can Taylor series be used to prove a difference involving logarithms?

[jj1986]

Homework Statement

Prove if t > 1 then log(t) - \int^{t+1}_{t}log(x) dx differs from -\frac{t}{2} by less than \frac{t^2}{6}

Homework Equations

Hint: Work out the integral using Taylor series for log(1+x) at the point 0

The Attempt at a Solution

Using substitution I get:
\int^{t+1}_{t}log(x) dx = \int^{t}_{t-1}log(u+1) du

The taylor series for log(1+x) = \sum (-1)^{n+1}x^{n}/n beginning at n=1.

So
\int^{t}_{t-1}log(u+1) du = \sum (-1)^{n+1}u^{n+1}/[(n)(n+1)]^{t}_{t-1} beginning at n=1.

I'm still not sure how to prove what the problem statement is asking me to prove. I see that the first term of the series is multiplied by 1/2 and the second term is multiplied by 1/6 and 1 degree higher so I'm sure that needs to be used somehow, I just don't see how exactly. Any suggestions?

 

[Billy Bob]

Are you sure the problem is stated correctly? log(x) can be integrated by parts (or look it up) without using series. Then graph to see if the question even makes sense.

 

[jj1986]

The problem is correct as stated and makes sense. I need to show that for any t > 1

| ( log(t) - \int^{t+1}_{t}log(x)dx ) - \frac{t}{2} | < \frac{t^2}{6}

Graphing it isn't really sufficient proof

 

[Billy Bob]

False when t=2.

 

[jj1986]

My mistake I misread the problem I need to show that for any t > 1

| ( log(t) - \int^{t+1}_{t}log(x)dx ) - \frac{1}{2t} | < \frac{1}{6t^{2}}

 

[jj1986]

Any suggestions?

 

[Billy Bob]

Since you are hoping to get t's in the denominator, I would try this. Integrate before converting to series (in spite of the hint). After integrating, then use properties of logs to rearrange (experiment with various ways). Try to get something where you could convert to series. Example: log(t) could be log( 1 + (t-1) ). Right idea, but not good enough, because you want t's in the denominator. So, if you can get log( (t+1)/t ) somehow, then you can write log( 1+ (1/t) ). Finally, apply the series with that. It's worth a try.