Suppose the series $\sum_{n=1}^\infty n a_n$ converges. Then I would like show (if it is always true):

$$\lim_{N \to \infty} N \sum_{n= N}^\infty a_n = 0.$$

My work:

I started with the condition $a_n > 0$ for all $n$. I know that $\sum_{n=1}^\infty na_n$ converges and therefore $\lim_{N \to \infty}\sum_{n=N}^\infty na_n = 0$.

Since $Na_n \leq na_n$ for $n \geq N$ it holds that $N\sum_{n=N}^\infty a_n \leq \sum_{n=N}^\infty n a_n $ and therefore

$$0 \leq \lim_{N \to \infty}N\sum_{n=N}^\infty a_n \leq \lim_{N \to \infty}\sum_{n=N}^\infty na_n = 0 $$

But for a general sequence$\{a_n\}$ that is not always or eventually nonnegative or nonpositive is this still true?

I suspect it is not but could not find a counterexample.

  • 1
    $\begingroup$ Summation by parts. $\endgroup$ – Jack D'Aurizio Jan 15 '19 at 22:00

The sign of $a_n$ is not relevant, although it is somewhat harder to prove the result.

Assume there exists a finite number $S$ such that $S_N = \sum_{n=1}^N n a_n \to S$ as $N \to \infty$. Using summation by parts we have

$$ N\sum_{n=N}^M a_n = N\sum_{n=N}^M n a_n \frac{1}{n} = N\left[\frac{S_M}{M} - \frac{S_{N-1}}{N} + \sum_{n=N}^{M-1} S_n\left(\frac{1}{n} - \frac{1}{n+1} \right)\right]$$

Taking the limit as $M \to \infty$ we get

$$N\sum_{n=N}^\infty a_n = -S_{N-1} + N\sum_{n=N}^{\infty} S_n\left(\frac{1}{n} - \frac{1}{n+1} \right)$$

Since $S- \epsilon < S_n < S+ \epsilon$ for sufficiently large $n$ , it can be shown that the limit of the sum on the RHS is $S$ and, thus,

$$\lim_{N\to \infty}N\sum_{n=N}^\infty a_n = -S + S = 0$$

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  • $\begingroup$ Thank you. So that I understand fully you are saying $ S- \epsilon < N \sum_{n=N}^\infty S_n(1/n - 1/(n+1)) <S+ \epsilon$ because $N \sum_{n=N}^\infty(1/n - 1/(n+1)) = N(1/N) = 1$ and $S- \epsilon < S_n < S + \epsilon$ for $n \geq N$? $\endgroup$ – SAS Jan 16 '19 at 0:11
  • $\begingroup$ @SAS: Yes -- that is correct. $\endgroup$ – RRL Jan 16 '19 at 4:35

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