Find $\lim_{n \to \infty} \sum_{k = 2n}^{10n} \frac{1}{k}$

Question

Find $$\lim_{n \to \infty} \sum_{k = 2n}^{10n} \frac{1}{k}$$

This is how I attempted to solve this:
We need to find the limit of this expression: $$(\frac{1}{2n} + \frac{1}{2n+1} + \dots + \frac{1}{10n})$$ Which is in fact $$H_{10n} - H_{2n-1}$$ And so the problem becomes $$\lim_{n \to \infty} (H_{10n} - H_{2n-1})$$ Now, I use the approximation that $n \to \infty \Rightarrow H_n \sim \ln(n)$ $$ \lim_{n \to \infty}(\ln(10n) - \ln(2n-1)) = \lim_{n \to \infty}(\ln \frac{10n}{2n-1}) = \ln(5) $$ Is it the correct way to solve this?

Answer 1

Using an approximation is usually dangerous, unless you can back it up. I'd gave it more of a thought. You can also use integrals to calculate this limit in the following way:

$$\lim\limits_{n \to \infty} \frac{1}{2n}+\dots+\frac{1}{10n} = \lim\limits_{n\to \infty} \frac{1}{n} \sum\limits_{i=2}^{9}\sum\limits_{k=0}^{n-1} \frac{1}{i+\frac{k}{n}} = \sum\limits_{i=2}^9 \int\limits_0^1 \frac{1}{x+i}dx $$ $$ = \sum\limits_{i=2}^9 (\ln(1+i)-\ln(i)) = \ln10-\ln2 = \ln5 $$

Answer 2

Your approximation $H_n\sim\ln n$ is not quite precise enough to prove this. But there's a tighter asymptotic with an error term: $$H_n=\ln n+\gamma+O(1/n)$$ which suffices. Here $\gamma$ is Euler's constant. Considering $H_{5n}-H_{2n-1}$ the $\gamma$s cancel, and the error term is still $O(1/n)$ which goes to zero.

Answer 3

Hint : Use Riemann sums.

$$\lim_{n\to \infty} \sum_{2n}^{10n} \frac 1k=\lim_{n\to \infty} \frac 1n \sum_{2n}^{10n} \frac nk=\int_{\frac {1}{10}}^{\frac 12} \frac {dx}{x}$$