Calculate $\lim_{n\to\infty} \frac{\sum_{i=1}^n ia_i}{n^2+n}$ given that $a_n \to a$

Question

Calculate $\displaystyle \lim_{n\to\infty} \frac{\sum_{i=1}^n ia_i}{n^2+n}$ given that $a_n \to a$.

I can see that if I can extract something from the sum the and have something multiplied by $\sum_i^n i = (n^2 +n)/2$ then stuff would cancel out, but I cant figure out what I can extract besides some arbitrary bound which won't converge.

Answer 1

Note that the converging sequence $\{|a_n-a|\}_n$ is bounded by some constant $M$. Moreover, for $\epsilon>0$, there is $N$ such that for $n> N$, $|a_n-a|<\epsilon$. Now recall that $\sum_{i=1}^n i=(n^2+n)/2$ and split the sum: for $n>N$ $$\frac{\sum_{i=1}^n ia_i}{n^2+n}-\frac{a}{2}=\frac{\sum_{i=1}^N i(a_i-a)}{n^2+n}+\frac{\sum_{i=N+1}^n i(a_i-a)}{n^2+n}.$$ Hence $$\left|\frac{\sum_{i=1}^n ia_i}{n^2+n}-\frac{a}{2}\right|\leq M\cdot\frac{\sum_{i=1}^N i}{n^2+n}+\epsilon\cdot\frac{\sum_{i=N+1}^n i}{n^2+n} \leq \frac{MN^2}{n^2+n}+\frac{\epsilon}{2}.$$ Can you take it from here?

Another way: use Stolz–Cesàro theorem.

Answer 2

As you mentioned, the idea behind this formula is: $$\sum_{i=1}^n i= \frac{n(n+1)}{2}$$

Suppose first that $a = 0$. You can write for $n > N \ge1$ $$\begin{aligned} \left\vert \frac{\sum_{i=1}^n ia_i}{n^2+n} \right \vert &\le \frac{\left\vert\sum_{i=1}^N ia_i \right\vert}{n^2+n} + \frac{\sum_{i=N+1}^n i \vert a_i \vert}{n^2+n}\\ &\le\frac{\left\vert\sum_{i=1}^N ia_i \right\vert}{n^2+n} + \frac{\sum_{i=1}^n i \vert a_i \vert}{n^2+n} \end{aligned}$$

Now for $\epsilon >0$, take $N$ large enough for having $\vert a_n \vert \le \epsilon/2$ for $n \ge N$. This is possible as we supposed $\lim\limits_{n \to \infty} a_n= 0$. Take then $M \ge N$ such that for $n \ge M$ you have $$\frac{\left\vert\sum_{i=1}^N ia_i \right\vert}{n^2+n} \le \epsilon/2$$.

This is possible as the numerator of the fraction is fixed, while the denominator tends to infinity. Finally for $n > M \ge N$ you get $$ \left\vert \frac{\sum_{i=1}^n ia_i}{n^2+n} \right \vert \le \frac{\left\vert\sum_{i=1}^N ia_i \right\vert}{n^2+n} + \frac{\sum_{i=N+1}^n i \vert a_i \vert}{n^2+n} \le \frac{\left\vert\sum_{i=1}^N ia_i \right\vert}{n^2+n} + \epsilon/2 \frac{\sum_{i=1}^n i }{n^2+n} \le \epsilon$$

Proving that $\lim_{n\to\infty} \frac{\sum_{i=1}^n ia_i}{n^2+n} = 0.$

Now for the general case notice that $$\frac{\sum_{i=1}^n ia_i}{n^2+n} - a/2 = \frac{\sum_{i=1}^n ia_i}{n^2+n} - \frac{\sum_{i=1}^n i a}{n^2+n}=\frac{\sum_{i=1}^n i(a_i-a)}{n^2+n}$$ to prove that $\lim_{n\to\infty} \frac{\sum_{i=1}^n ia_i}{n^2+n}=a/2$ using the case $a=0$.