# Show that Euler-Mascheroni sequence is decreasing &monotonic

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1. Jan 31, 2015

### timnswede

1. The problem statement, all variables and given/known data
tn=1+1/2+1/3+...+1/n - ln(n)
a.) Interpret tn - tn+1= [ln(n+1)-ln(n)] - 1/(n+1) as a difference of areas to show that tn - tn+1 > 0.

2. Relevant equations

3. The attempt at a solution
I have not started working on part b) yet, because so far I am stuck on part a). I just simplified a bit and got ln(1+1/n)>1/(n+1). Not sure how to prove that the left side is bigger than the right.

2. Jan 31, 2015

### Staff: Mentor

Did you consider an integral?
There are other methods, too, depending on the equations you got for the logarithm.

3. Jan 31, 2015

### timnswede

Is it as easy as just integrating both sides? I get xln(1+1/x)+ln(1+x)>ln(1+x). The left side is always greater since xln(1+1/x) is always greater than zero for n>1. Is that enough to prove it?

4. Jan 31, 2015

### Staff: Mentor

Integrate both sides with respect to what?

[ln(n+1)-ln(n)] looks like an integration result.

5. Jan 31, 2015

### timnswede

Woops those x's up there should be n's. But with dn. I can't see what integral would have resulted in [ln(n+1)-ln(n)] though. Integrating 1/n gets me ln(n), but integrating ln(n) does not get me ln(n+1).

6. Jan 31, 2015

### Staff: Mentor

If you integrate from where to where?

7. Jan 31, 2015

### timnswede

From the beginning of the sequence, 1, to the end, n.

8. Jan 31, 2015

### utkarsh009

What if you integrate it from n to n+1?
Also can you think of the area of rectangle formed with height 1/(n+1) and width (n+1)-n? Just compare them on the graph of f(x)=1/x