Calculating Limits and Harmonic Series: A Straightforward Approach

[kotreny]

I don't know where to begin to calculate this limit, but I think I know what the answer is. Could someone please do it with straightforward methods? I suspect the answer is ln(x+1).
Also, I don't know how to write it down in proper notation, so I ask that someone do that too.

lim_dx-->0 SUM_{k=1 to x/dx}[dx/(1+(k-1)dx)]

 

[Some Pig]

\lim_{\Delta x\to0}\sum_{k=1}^{[\frac{x}{\Delta x}]}\frac{\Delta x}{1+(k-1)\Delta x}\\<br /> =\lim_{n\to\infty}\sum_{k=1}^n\frac{x/n}{1+(k-1)x/n}\\<br /> =\lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac{x}{1+x(k-1)/n}\\<br /> =\int_0^1\frac{x}{1+xu}\ du<br /> =\log(1+x)

 

[kotreny]

I can't believe some pig answered my request! Ugh! No seriously, thanks. (Hey, you asked for it.)

The thing that interests me about this formula is that if you replace delta x with 1 instead of -->0, you get the harmonic series up to the x-th term. And you can generalize it by having delta x be any number. The natural log is simply the case where delta x -->0, or in other words, this thing is "compounded continuously." It's an expression of the intuited difference between the harmonic series at x and the natural log: Discrete vs. Continuous.

 

[HallsofIvy]

No, he did not replace \Delta x with 1, he replaced it with x/n which does go to 0 as n goes to infinity.

 

[kotreny]

I know, and the limit evaluates to ln(1+x). I'm saying if we replace delta x in the first expression with 1, we get the harmonic series summed up until k=x.