# Proving convergence of a two related sequences, specifically that one converges to a value half of the other

Suppose a sequence $$a_n$$ converges to some $$x \in \mathbb{R}$$.

Let $$b_n$$ be defined by $$b_n = \frac{1}{n^2}(a_1 + 2a_2 + ... +na_n)$$. Prove that $$b_n$$ converges to $$\frac{x}{2}$$.

I am confused by the above prompt in my textbook, I have looked at some proofs and I see how to get that $$c_n = \frac{1}{n}(a_1 + a_2 + ... +a_n) = x$$, but I am struggling to see how to expand that given the coefficients in $$b_n$$.

Any and all help is appreciated.

• The question does not specify if it is monotonic!
– user732172
Commented Apr 5, 2020 at 18:16

Let $$a_n=x+d_n$$, then eventually $$|d_n|\lt\epsilon$$.
$$b_n-\frac1{n^2}\frac{n^2+n}2x = \frac1{n^2}(d_1+\ldots+nd_n)$$ Now divide the sum into a finite first part $$k\lt N$$ where $$d_k$$ may be large, and the rest where all the $$d_k$$ have absolute value below $$\epsilon$$.
• I'm confused on how to use the equation you've given, as I don't see it was found from $b_n - \frac{x}{2}$, which I thought was what I needed to show was less than $\epsilon$
• I subtracted $\frac1{n^2}(x+2x+\ldots+nx)$ from both sides Commented Apr 5, 2020 at 18:24
• Sorry -- still not fully understanding the logic behind it. I'll ask some more clarifying questions then. So you have $b_n - \frac{x}{2} = \frac{1}{n^2}(a_1 +... +na+n)$ and then you subtract $b_n$ from both sides? I'm still very confused as to what you are subtracting from... or what \$\frac{(n^2+n)x}{n^2} is...