4

Suppose $a_n$ is a nonnegative real sequence such that \begin{equation} \sum_n a_n <\infty. \end{equation}

What do we know about $a_n$? We know $a_n\to 0$. We know $$\lim\inf n a_n = 0.$$ But can we say $$\lim n a_n = 0?$$ If not, why not? That is, can you construct a sequence that is summable but this is not the case?

Martin Sleziak
  • 53,687
Patrick Johnstone
  • 105

3 Answers3

4

If $n$ is a perfect square, let $a_n=\frac{1}{n}$. If $n$ is not a perfect square, let $a_n=\frac{1}{2^n}$.

The series $\sum a_n$ converges, but $na_n=1$ whenever $n$ is a perfect square. So it is not true that $\lim_{n\to\infty} na_n=0$.

The idea is that if we make $a_n=\frac{1}{n}$ (kind of big) on a sparse enough set, with the rest of the $a_n$ "small," we will have convergence.

André Nicolas
  • 507,029
  • And more generally, for any sequence $b_n \to 0$ there are $a_n$ with $\sum_n a_n$ convergent that share infinitely many elements with $b_n$. So there is nothing we can say of the form $g(n, a_n) \to L$ that isn't already implied by $a_n \to 0$. – Robert Israel Mar 30 '16 at 23:34
2

The answer is no. More strongly the following is true.

Let $b_n$ be any sequence whose limit is 0. Then there is a sequence $a_n$ which is summable, and is equal to $b_n$ for infinitely many $n$.

Proof. Let $a_n = b_n$ the first time that $|b_n| < 1$, the first time after that that $|b_n| < \frac{1}{2}$, the first time after that that $|b_n| < \frac{1}{4}$ and so on. At all other points let $a_n$ be 0. By construction it is equal to $b_n$ infinitely often, yet the sum of the absolute value of its terms is bounded above by $1+\frac{1}{2} + \frac{1}{4} + ... = 2$.

Now apply this result to $b_n = \frac{1}{n}$. Or, more pathologically, $\frac{1}{\sqrt{n}}$. Or any other o(1) sequence that you want.

btilly
  • 1,186
1

As the other anwsers have stated, this is not true in the general case. However, if you suppose $(a_n)$ non increasing, then you have $\lim\limits_{n \to \infty} n \cdot a_n = 0 $.

To see why, consider $\sum\limits_{k=n+1}^{2n} a_k$.

In one hand, $\sum\limits_{k=n+1}^{2n} a_k \ge \sum\limits_{k=n+1}^{2n} a_{2n} = n \cdot a_{2n} $.

On the other hand, $\sum\limits_{k=n+1}^{2n} a_k = \sum\limits_{k=0}^{\infty} a_k - \sum\limits_{k=0}^{n} a_k- \sum\limits_{k=2n+1}^{\infty} a_k \to 0 $ when $ n \to \infty$.

Therefore $\lim\limits_{n \to \infty} 2n \cdot a_{2n} = 0$.

And finally $(2n+1) \cdot a_{2n+1} \le (2n+1) \cdot a_{2n} = 2n \cdot a_{2n} + a_{2n} \to 0 $ when $ n \to \infty$.

yultan
  • 1,417