Pulling sigma notation out of thin air The function $ f(t) $ satisfies $ \, 0 \leq f(t) \leq k $ for $ \, 0 \leq t \leq 1 $, where $ k $ is a constant.
Therefore, $$ 0 \leq \int_{0}^{1}f(t)\, dt \leq k $$
So, if $$ f(t) = \frac{1}{n-t}\, -\, \frac{1}{n} $$
then if $ n > 1 $ $$ 0 \leq \ln \left ( \frac{n}{n-1} \right ) - \frac{1}{n} \leq \frac{1}{n-1} - \frac{1}{n} $$
What I need to do is hence show that $$ 0 \leq \ln 2 - \sum_{n = N+1}^{2N}\frac{1}{n} \leq \frac{1}{2N} $$
I'm unsure how to do this. I can show it for the specific case where $ N = 1 $, obviously, but I'm pretty sure my teacher wants it shown more generally. Where do I get the summation from?
Thanks!

Consider the sum 
$$S(N) = \sum_{n=N+1}^{2N} \int_0^1 f(t) dt.$$ 
Applying your integral estimate to each summand gives
$$0 \le \sum_{n=N+1}^{2N}
\ln \left( \frac{n}{n-1} \right) - 
 \sum_{n=N+1}^{2N}\frac{1}{n} 
 \le \sum_{n=N+1}^{2N} \left( \frac{1}{n-1} - \frac{1}{n} \right).$$
Then apply $\ln(a b) = \ln(a) + \ln(b)$ to the first sum to get
$$ 0 \le  \sum_{n=N+1}^{2N} \left( \ln (n) - \ln (n-1) \right) - \sum_{n = N+1}^{2N}\frac{1}{n} \le\sum_{n=N+1}^{2N} \left( \frac{1}{n-1} - \frac{1}{n} \right).$$
You should notice we have two telescoping sums here - the first down to $$\ln(2N)-\ln(N)=\ln 2$$ and the last down to $$\frac{1}{N} - \frac{1}{2N} = \frac{1}{2N},$$ giving you the desired inequality.
