Question:
Show that:
$$\lim_{n \to \infty} \frac{1}{\ln n} \sum_{k=1}^{n} \frac{1}{2k-1} = \frac{1}{2}$$
Attempted answer:
My thinking is that we should close in the sum with the help of two integrals. Hopefully this will resolve the entire expression as the left part goes towards zero as n goes to infinity and both come out as $1/2$
What integrals should we use? I am thinking:
$$ \int_{1}^{n} \frac{1}{2x} \leq \sum_{k=1}^{n} \frac{1}{2k-1} \leq \int_{1}^{n} \frac{1}{x}$$
Let us evaluate the two integrals:
$$ \int_{1}^{n} \frac{1}{2x} = \frac{\ln{x}}{2} + C_{1}$$
$$ \int_{1}^{n} \frac{1}{x} = \ln{x} + C_{2}$$
So regardless of the situation, we can get rid of the $\ln{n}$ in the denominator from the original expression.
Using the first integral, the expression evaluates to $1/2$, and the second to $1$.
However, this does not persuade me that the answer is really $1/2$. After all, it could be anywhere between the two (inclusive).
How to complete the solution to this question? Is it better to choose integral estimates of the sum instead? If so, which should be used?
Best Answer
Create your integral bounds so that the sum you gave will be an upper Riemann sum for the lower bound and a lower Riemann sum for the upper bound. You did that for the lower bound case, for the upper bound case: $$\sum_{k=1}^\infty \frac{1}{2k-1}= 1 + \sum_{k=1}^\infty \frac{1}{2k+1}$$ Now you can think of $\sum_{k=1}^n \frac{1}{2k+1}$ as the lower Riemann sum of $\frac{1}{2x+1}$ from $x=0$ to $n$. So $\sum_{k=1}^n \frac{1}{2k+1} \leq \int_0^n \frac{1}{2x+1}$. Thus: $$\sum_{k=1}^n \frac{1}{2k-1} \leq 1 + \frac{\ln(2n+1)}{2}$$ So showing that $\frac{\ln(2n+1)}{\ln(n)}$ approaches $1$ will suffice. And for that, notice $\frac{\ln(2n+1) - \ln(2n)}{\ln(n)}$ approaches $0$ and $\frac{\ln(2n)}{ln(n)} = \frac{\ln(2) + \ln(n)}{\ln(n)}$ approaches $1$. Thus $\frac{\ln(2n+1)}{\ln(n)} = \frac{\ln(2n+1) - \ln(2n)}{\ln(n)} + \frac{\ln(2n)}{\ln(n)}$ approaches $1$.