The limit of 2 sequences

Question

I have encountered this problem in an olympiad magazine:

Let $a \in \mathbb{R}$ and
$$x_n = 1 - \frac{1}{2^a} + \frac{1}{3^a} - \dots + \frac{1}{(2n - 1)^a} - \frac{1}{(2n)^a}$$ $$y_n = \frac{1}{(n+1)^a} + \frac{1}{(n+2)^a} + \dots \frac{1}{(2n)^a}$$

Compute $\lim_{n \to \infty} \frac{x_n}{y_n}$

I tried to get a relation between $x_n$ and $y_n$, but to no avail. Any other ideas?

Answer 1

We have $1-\frac{1}{2^a} < x_n < 1$ and $\frac{n}{(2n)^a}\le y_n \le \frac{n}{n^a}$.

• When $a < 1$ then $0 \le \frac{x_n}{y_n} \le 2^a n^{a-1} \to 0$.

• When $a > 1$ then $\frac{x_n}{y_n} \ge (1-\frac{1}{2^a}) n^{a-1} \to \infty$.

• When $a=1$, then

  • $x_n \to \log 2$ (plug $x=1$ into the Taylor series of $\log(1+x)$), and
  • $y_n = \sum_{k=n+1}^{2n} \frac{1}{k}$ which can be sandwiched between $\int_n^{2n+1} \frac{dx}{x} = \log\frac{2n+1}{n} \to \log 2$ and $\int_{n-1}^{2n} \frac{dx}{x} = \log \frac{2n}{n-1} \to \log 2$.