Question

if Sn is the nth partial sum of the harmonic series, show that e^Sn > n+1. Why does this imply that the harmonic series is divergent?

Answer 1

let me get my old review books..

Answer 2

I suppose ln'ing it tells you something...

Why do you like this stuff, hokiefan?

Answer 3

let's try lning actually i didn't think of it:
Sn > Ln(n+1).. hmm now what..?

Answer 4

i dont really like it haha i just have to do it because i have a test coming up. i hate series and sequences.

Answer 5

Yup, me too:-)

Answer 6

I know.. sorry I can't help but yea series and sequences were the worst part of my calc stuff..i regret never learning them properly... i needed them for diff eqns.. so i had to spend hours studying those in addition... but then i just studied fourier series..

Answer 7

dam ok thanks for trying guys

Answer 8

try yahoo answers.. those guys are usually smarter than us..my questions rarely get answered here too.. well unless polpak is on..

Answer 9

Let's see, the partial sums have logarithmic growth and we have an equation with ln...

Answer 10

Okay hold up I got an idea.. if the limit of the partial sums is some number, then the series converge right? So, if we take Lim n->infinity (Sn) it will be greater than
Limit n->infinity ln(n+1)

Answer 11

So limit as n->infinity of Ln(n+1) is infinity.. and what's greater than infinity? Infinity.. so Sn doesn't converge either.

Answer 12

I am 75-80% sure that this is how u do it..

Answer 13

For Sn 1 to k it's ln k + a constant...

Answer 14

doesnt matter.. as k approaches infinity.. Lninfinity is undefined..

Answer 15

The n th partial - ln n converges to a constant..

Answer 16

oh i see what you mean..

Answer 17

Close to it now, let me think for a bit...

Answer 18

The second part follows immediately since as was mentioned above as n tends to inf, so does ln(n+1) and we have something larger still hence divergent. The first part is really quite tricky.

Answer 19

I've got it, hold on while I try to explain...

Answer 20

We have \[s_n=\sum_{k=1}^{n}(1/k)>\int\limits_{1}^{n+1}(1/x)dx=\ln(n+1)\] So take the exponential and you're done. The inequality works because if you imagine the curve 1/x and then the sn are approximating the area underneath this from above using rectangles of width 1 just touching the curve, whereas the defn of the integral is the area underneath. Make sense?

Answer 21

If you think about 1/x and the rectangle method way of integration (area under a curve) with rectangles 1*1/n

1+1/2+1/3... etc for the rectangles.

And we know that int 1 to inf of 1/x is infinite and this area is within the rectangles.

So sum 1 to k of 1/n is > int 1 to k+1 of 1/x = ln(k+1).

Answer 22

@makaze Snap, yours is prettier though...

Answer 23

Haha, within seconds!

Answer 24

Thanks everyone for your help, i really appreciate it.