Well, what you have looks good. The 1/n becomes "dx", and you can write n/(n+i) as 1/(1+i/n), and the i/n becomes "x", so perhaps you get something like [math] \int_0^1 1/(1+x)\, dx = \log(2) [/math]

>>26956
Thank you, that's what I did.

>>26939
>harmonic numbers
That's what the question is asking, but it's not the intended solution.

I think that the question can be solved using the integral of 1/x.

You will find the summands as a Riemann-sum of the function 1/x in the interval [n;2n+1] using unit-length intervals and function values at the left margin. This sum is greater than the integral of (1/x) between n and 2n+1.

Since int (1/x) dx = ln(x) + c you obtain that the sum is greater than ln(2n+1) - ln(n) > ln(2n) - ln(n) = ln(2) + ln(n) - ln(n) = ln(2).

The limes is therefore greater or equal to ln(2).

The second inequality can perhaps be done similarly taking a Rieman sum that is smaller than the integral of 1/x.

>>26974
too late

>>26974
useful none the less

>>26974
I suppose you could do the same bounding procedure using 1/(1+x) as well, to make the whole process more rigorous.