# Convergence of series implies convergence of related series

I am considering a problem in Browder's Mathematical Analysis:

Let $(a_n)_{n=1}^{\infty}$ be a sequence in $\mathbb{R}$.

(a) Show that if $\lim_{n \rightarrow \infty} a_n = A$ exists, then $$\lim_{n \rightarrow \infty} \frac{a_1 + \dots + a_n}{n} = A.$$

Proof: Without loss of generality, suppose $a_n \rightarrow 0$. Fix $\epsilon > 0$. Then there exists an $N \in \mathbb{N}:\ \forall n \geq N$, $|a_n| < \epsilon.$ Set $S_N = \sum_{k=1}^N |a_k|.$ Choose $\tilde{N} > N:\ \forall n \geq \tilde{N}$, $$\frac{S_N}{n} < \epsilon.$$ Then if $n \geq \tilde{N} > N$, \begin{align*} \left|\frac{a_1 + \dots + a_n}{n}\right| & \leq \frac{1}{n}\left(|a_1| + \dots + |a_n|\right) \\ & = \frac{S_N}{n} + \frac{|a_{N+1}| + \dots + |a_n|}{n} \\ & < \epsilon + \frac{(n-(N+1))\epsilon}{n} \\ & \leq 2 \epsilon. \end{align*}

(b) Show that if $\sum a_n$ converges, then $$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n ka_k = 0.$$

Unfortunately I am having trouble proving part (b). I believe that the assumption that the series $\sum a_n$ converges tells us that the sequence $na_n \rightarrow 0$, i.e. that the sequence $a_n$ converges fairly quickly to zero. If this is the case, I can invoke the result in part (a) and conclude but I am not sure how to proceed. I've seen an extremely similar question asked here: Series converges implies $\lim{n a_n} = 0$ but this question does not place any assumptions on the monotonicity on the sequence $a_n$.

• I think that part b) can be solved with summation by parts (not totally sure). The derived sequence of part a) is named Césaro sum. – Masacroso Apr 30 '17 at 0:28

For part (b), assume $S_n = \sum_{j=1}^n a_j \to S$ as $n \to \infty.$

We have

$$\sum_{j=1}^nS_j = \sum_{j=1}^n\sum_{k=1}^ja_k = \sum_{j=1}^n\sum_{k=1}^na_k 1_{(k \leqslant j)}= \sum_{k=1}^n\sum_{j=1}^na_k 1_{(k \leqslant j)}$$

where we use the indicator $1_{(k \leqslant j)} = 1 (\,\,\text{if} \,\,k \leqslant j), = 0 (\,\,\text{if} \,\,k > j)$ and switch the sums.

Thus

$$\sum_{j=1}^nS_j = \sum_{k=1}^n\sum_{j=1}^na_k 1_{(k \leqslant j)} = \sum_{k=1}^n\sum_{j=k}^na_k = \sum_{k=1}^n a_k (n - k +1) \\ = (n+1) \sum_{k=1}^n a_k - \sum_{k=1}^n k a_k,$$

and

$$\tag{*}\frac{1}{n} \sum_{k=1}^n k a_k = (1 + 1/n)S_n - \frac{1}{n} \sum_{j=1}^n S_j.$$

Since $S_n \to S$ it follows that $\frac{1}{n} \sum_{j=1}^n S_j \to S$ by the result of part (a), and taking the limit of (*) we get

$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n k a_k = 1 \cdot S - S = 0.$$

Alternatively, summing by parts with $S_0 := 0$ we get

\begin{align}\sum_{k=1}^n k a_k &= \sum_{k=1}^n k(S_k - S_{k-1}) \\ &= nS_n+ \sum_{k=1}^{n-1}k S_k - \sum_{k=2}^nkS_{k-1} \\ &= nS_n + \sum_{k=1}^{n-1}k S_k - \sum_{k=1}^{n-1}(k+1)S_{k} \\ &= nS_n - \sum_{k=1}^{n-1} S_k \\ &= (n+1)S_n - \sum_{k=1}^n S_k \end{align}

Now proceed as before.

• This is very slick thank you! – xk3 Apr 30 '17 at 2:19
• You're welcome. The summation by parts approach that I added is the typical way to go, but the trick in the first approach should prove useful for other types of problems. – RRL Apr 30 '17 at 2:22