$\sum_{k=1}^{n} \frac{1}{n+k} < \ln 2$

Question

$\sum_{k=1}^{n} \frac{1}{n+k} < \ln 2$

Here are some proofs for this inequality but I have another approach. Can anyone verify my proof? Thanks.

$$\sum_{k=1}^{n} \frac{1}{n+k}< \int_0^n \frac{1}{n+x}dx=\ln (n+x)\mid ^n_0 = \ln(2n)-\ln(n) = \ln 2.$$

Does the first inequality of this proof hold? In general, what's the relationship between a series and its integral form? Is the series always less than its integral form? Is it the same for a finite summation?

Answer 1

Here is a (French, but efficient) way of understanding how the comparison series/integrals work for decreasing functions. In your case $$\sum_{k=1}^{n} \frac{1}{n+k}< \int_1^{n+1} \frac{1}{n+x-1}dx=\ln (n+x-1)\mid ^{n+1}_1 = \ln(2n)-\ln(n) = \ln 2.$$