5

I have the series $\sum a_n$ where

$$a_n = \begin{cases} \frac{1}{n}, &\text{$n$ is a perfect square} \\ \frac{1}{n^2}, &\text{otherwise}\end{cases}$$

Prove that $\sum_{n=1}^\infty R_n$ converges where remainder $R_n = \sum_{i=n}^\infty a_i$.

My work: I can show that the series $\sum a_n$ converges because

$$\sum_{n=1}^Na_n = \sum_{n \neq k^2} \frac{1}{n^2} + \sum_{n = k^2, 1 \leq k^2 \leq N} \frac{1}{k^2} \leq 2 \sum_{n=1}^N \frac{1}{n^2}$$

and this means the remainders converge to $0$, $\lim_{n\to \infty}R_n = 0$. Hence, I can't rule out convergence of $\sum R_n$ by the term divergence test. But I can't find a comparison to prove it converges.

Michael Hardy
  • 1
scobaco
  • 597

2 Answers2

3

Actually, the remainder series does not converge in this case.

Assume that $\sum_{n=1}^\infty R_n$ is convergent. The sequence $R_n$ converges to $0$, as you showed, but it is also monotonically decreasing since $R_n - R_{n+1} = a_n > 0$ for all $n$. As is well-known for a convergent series with monotonically decreasing terms, $\lim_{n \to 0} n R_n = 0.$

However, this implies that as $n \to \infty$,

$$n a_n = nR_n - nR_{n+1} = nR_n - \frac{n}{n+1}(n+1)R_{n+1} \to 0,$$

which is impossible since $\limsup na_n = 1 \neq 0$.

RRL
  • 90,707
2

Another possibility is to focus on the perfect square terms so $$R_n>\sum_{i=\lceil\sqrt n\rceil}^{\infty}\frac1{i^2}>\int_{\sqrt n+1}^{\infty}\frac1{i^2}di=\left.-\frac1i\right|_{\sqrt n+1}^{\infty}=\frac1{\sqrt n+1}$$ Then $$\sum_{n=1}^{\infty}R_n$$ Diverges by direct comparison with $$\sum_{n=1}^{\infty}\frac1{2\sqrt n}$$ The same approach could show that the series diverges even without the perfect square terms being larger non the corresponding non-square terms.

user5713492
  • 15,938