# Asymptotic behavior of series tail

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.

• Summation by parts. – 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,

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

• 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 $n \geq N$? – SAS Jan 16 '19 at 0:11
• @SAS: Yes -- that is correct. – RRL Jan 16 '19 at 4:35