Q:

how to bound harmonic numbers?

Consider the following sum,
$$
\sum_{x=1}^{L} \frac{1}{cx + 1} 
$$
Does it grow as the logarithm of $L$? Are there any good upper and lower bounds for this expression as $L$ is large enough? I would like the bounds to depend on $c$ and on $L$.

A:

If $c=0$, the sum is simply $L$. If $c>0$ you can compare with $\int_0^{L}\frac{\mathrm dx}{cx+1}$ and $\int_1^{L+1}\frac{\mathrm dx}{cx+1}$, i.e. we have
$$ \tag1\frac1c\ln(cL+1)>\sum_{k=1}^L\frac1{ck+1}>\frac1c\ln(cL+c+1)-\frac1c\ln(c+1).$$ 
In short: The sum is $\approx\frac1c\ln L$ for large $L$.
The difference between left and right in the simple exstimate (1) is merely $\frac1c\ln\frac{(cL+1)(c+1)}{cL+c+1}\to \frac1c\ln(c+1)$, i.e. bounded by a constant.
If $c<0$, treat the first few summands "manually" until $cx+1>0$ and then proceed as in the positive case.

