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.
A: We see that for every $c>1$,  $cx<cx+1<c(x+1)$ so $$\frac{1}{c(x+1)}<\frac{1}{cx+1}<\frac{1}{cx}$$
which means that we can find some bounds for the sum you want from the 2 sums:
$$\sum{\frac{1}{c(x+1)}}<\sum{\frac{1}{cx+1}}<\sum{\frac{1}{cx}}$$
We know that $$\sum{\frac{1}{c(x+1)}}=\frac{1}{c}\sum{\frac{1}{x+1}}\sim\frac{1}{c}\cdot lnL$$ and   $$\sum{\frac{1}{cx}}=\frac{1}{c}\sum{\frac{1}{x}}\sim\frac{1}{c}lnL$$
because it is well known that $H_n=\sum_{x=1}^{L}\frac{1}{x}\sim lnL$
So the sum you want is $\sim \frac{1}{c}lnL$ and the bounds you want are mentioned above.
The case  $c\leq1$ can be investigated in the same way.
