>>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.
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