I've always had trouble with sequences and series, and I'm getting ready for upcoming finals now.

There's an example in my calculus text that says:

Show that the harmonic series $$\sum$$1/n is divergent.

The solution states: For this particular series it's convenient to consider the partial sums s2,s4,s8,s16,s32,...and show that they become large.

s1=1
s2=1 + (1/2)
s4= 1 + 1/2 + (1/3 + 1/4) > 1+ 1/2 + (1/4 + 1/4) = 1+ 2/2
s8=1 + 1/2 + (1/3 + 1/4) +(1/5 + 1/6 + 1/7 + 1/8) > 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) = 1 +1/2 +1/2 +1/2 = 1+3/2

so the pattern becomes s2n > 1+ n/2
which shows that s2n -> infinity as n -> to infinity and so {sn} is divergent. Therefore the harmonic series diverges.

What I don't understand is why the terms get substituted, for smaller ones (1/3+1/4 becomes 1/4+1/4, etc). If I understood why they were doing that I would understand the rest of it, but there's no explanation in the book, unless it was covered in an earlier section.

Any help understanding would be most appreciated.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Recognitions: Gold Member Science Advisor Staff Emeritus Because you want ">". If S= a+ b+ c and I substitute x>a for a, y> b for b, z> c for c, then obviously S< x+ y+ z.
 Recognitions: Homework Help Smaller terms are sustituted for simplicity. The problem with sums is that it is often hard to evaluate a particular sum. Our goal hear is to show a sum diverges. It suffices to show the sum excedes any particular value. We want to show the sum is big, but it is hard to actually find out how big (log(n)). It is easier to show it is at least as big as a simpler sum. If we have two sums the one with larger terms will be larger. By comparing the sum in question to a smaller easier one we have demostrated the sum is a least a big as the simple one. Say each of us has 100 numbered bags of nickels, and for any particular number your bag has more nickles than mine. We can concluded that you have more nickles total than I without counting either bag. We can also count the nickles in my bags and we have a lower bound on your nickles. In other words if I can afford to to buy a red wagon with my nickels, your will have at least enough to buy the same wagon (assum it offered at the same price).