Mean of two numbers by infinite sequences

Question

Consider two numbers $a$ and $b$, and the following sequence alternating between even and odd positions: $$ a+2b+3a+4b+5a+6b\ldots, $$ If we ''normalize'' $$ \frac{a+2b+3a+4b+\ldots}{1+2+3+4+\ldots}, $$ it turns out this ratio approaches the mean value of $a$ and $b$: $(a+b)/2$. In general $$ \frac{a+2^n b+3^n a+4^n b+\ldots}{1^n+2^n+3^n+4^n+\ldots}=\frac{a+b}{2} $$ for $n\geq1$. However if we use exponential functions instead powers: $$ \frac{m^1 a+m^2 b+m^3 a+m^4 b+\ldots}{m^1 +m^2 +m^3 +m^4 +\ldots} $$ for some $m>1$, this ratio oscillates and does not approach any number.

Could someone explain why convergence to the mean is obtained by using sequences of powers, and the ratio diverges for sequences of exponentials?

Answer 1

To give an evaluation, to make more rigourous, in the first case we have

$$1^n+2^n+3^n+4^n+\ldots+k^n\sim \frac{k^{n+1}}{n+1}$$

$$a+2^n b+3^n a+4^n b+\ldots+k^nb\sim a\left(\frac{k^{n+1}}{n+1}-2^n\frac{k^{n+1}}{(n+1)2^{n+1}}\right)+2^nb\frac{k^{n+1}}{(n+1)2^{n+1}}=$$$$=\frac12(a+b)\frac{k^{n+1}}{n+1}$$

and therefore

$$\frac{a+2^n b+3^n a+4^n b+\ldots+k^nb}{1^n+2^n+3^n+4^n+\ldots+k^n}\sim \frac{\frac12(a+b)\frac{k^{n+1}}{n+1}}{\frac{k^{n+1}}{n+1}}\to \frac{a+b}{2}$$

On the other hand for $m\neq 1$

$$m^1 +m^2 +m^3 +m^4 +\ldots+m^k =\frac{m^{k+1}-m}{m-1}$$

$$m^1 a+m^2 b+m^3 a+m^4 b+\ldots+m^kb\sim a\left(\frac{m^{k}-m}{m-1}-\frac{m^{k}-m^2}{m^2-1}\right)+b\frac{m^{k+2}-m^2}{m^2-1}=$$

$$=a\frac{m^{k+1}-m}{m^2-1}+b\frac{m^{k+2}-m^2}{m^2-1}$$

and therefore

$$\frac{m^1 a+m^2 b+m^3 a+m^4 b+\ldots+m^kb}{m^1 +m^2 +m^3 +m^4 +\ldots+m^k}\sim \frac {a+bm} {m+1}$$

but for

$$m^1 a+m^2 b+m^3 a+m^4 b+\ldots+m^ka$$

we would obtain

$$\frac{m^1 a+m^2 b+m^3 a+m^4 b+\ldots+m^ka}{m^1 +m^2 +m^3 +m^4 +\ldots+m^k}\sim \frac {am+b} {m+1}$$

Answer 2

Some thoughts; not sure if this fully answers "why" but an attempt:

The difference in coefficients of $a$ and $b$ goes roughly like the last coefficient term (whichever one's ahead is ahead by around that much). In the polynomial case this difference becomes small relative to the sums; in the geometric case, this stays large.

This is because last term of a geometric sequence with ratio $m > 1$, no matter how long, is a constant fraction of its sum. But for a polynomial sequence, eventually the last term becomes a vanishingly small fraction of the sum.

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More formally: consider a sequence of positive reals $c_1, c_2, \dots$ (may be geometric, polynomial, whatever). We're interested in

$$\frac{c_1a + c_2b + c_3a + \dots}{c_1 + c_2 + c_3 + \dots}$$

Or more precisely, let the partial sum $s_k = \sum_{1 \le i \le k} c_i$. Let the sum of odd terms $o_k = \sum_{1 \le 2i-1 \le k} c_{2i-1}$ and the sum of even terms $e_k = \sum_{1 \le 2i \le k} c_{2i} = s_k - o_k$. We're interested in

$$\lim_{k \rightarrow \infty} \frac{o_k a + e_k b}{s_k}$$

Now we can observe this converges for all $a, b$ if and only if $\frac{o_k}{s_k}$ converges.

So the question is: for what kinds of sequences does $\frac{o_k}{s_k}$ converge?

Don't have a nice full answer to this.

But we're more interested in the case when $c_i$ increases it seems, so let's restrict ourselves to that case. Hereafter $c_i$ are increasing.

In this case, we can say the following (the formal version of the above)

Claim:

(i) if $\frac{c_k}{s_k} \rightarrow 0$ then $\frac{o_k}{s_k} \rightarrow \frac{1}{2}$.

(ii) if $\frac{o_k}{s_k} \rightarrow L$ then $L = \frac{1}{2}$ and $\frac{c_k}{s_k} \rightarrow 0$.

Proof of claim:

Observe first that $\frac{o_k}{s_k} \rightarrow L$ if and only if $\frac{o_k - e_k}{s_k} \rightarrow 2L-1$. In particular $\frac{o_k}{s_k} \rightarrow \frac{1}{2} \Leftrightarrow \frac{o_k - e_k}{s_k} \rightarrow 0$

(i) Assume $\frac{c_k}{s_k} \rightarrow 0$. Because the $c_i$ are increasing and positive: when $k$ is odd, $0 \le o_k - e_k \le c_k$. When $k$ is even $0 \le e_k - o_k \le c_k$. So $\left|\frac{o_k - e_k}{s_k}\right| \le \frac{c_k}{s_k}$ for all $k$, hence $\left|\frac{o_k - e_k}{s_k}\right| \rightarrow 0$.

(ii) Assume $\frac{o_k}{s_k} \rightarrow L$. We have $\frac{o_k - e_k}{s_k} \rightarrow 2L-1$. But as in (i) we have $o_k - e_k \ge 0$ for odd $k$ and $o_k - e_k \le 0$ for even $k$. So we must have $2L-1 = 0 \Rightarrow L = 1/2$.

Now let's try to upper bound $\frac{c_k}{s_k}$ in terms of $|\frac{o_k - e_k}{s_k}|$.

For any odd $k$, $(e_{k-1} - o_{k-1}) + (o_k - e_k) = o_k - o_{k-1} = c_k$. Both those terms are positive, so we have that either $\frac{c_k}{2} \le e_{k-1} - o_{k-1} = |o_{k-1} - e_{k-1}|$ or $\frac{c_k}{2} \le o_k - e_k = |o_k - e_k|$. This also holds for even $k$ and can be shown similarly.

So we have

$$c_k \le 2\left(\max \{|o_{k-1} - e_{k-1}|, |o_k - e_k|\}\right)$$

Divide both sides by $s_k$, and we see that the RHS goes to 0 and hence $\frac{c_k}{s_k} \rightarrow 0$ $\square$.

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Applied to polynomial case: so we have $c_1, c_2, \dots = 1^n, 2^n, \dots$. Let's lower bound $s_k$ by some multiple of $k^{n+1}$. Very roughly, note that for say $k \ge 4$, $$s_k = \sum_{i=1}^k i^n \ge \sum_{i \ge k/2} i^n \ge \sum_{i \ge k/2} \left(\frac{k}{2}\right)^n \ge \lfloor \frac{k}{2} \rfloor \left(\frac{k}{2}\right)^n \ge \frac{k}{4}\left(\frac{k}{2}\right)^n = \frac{k^{n+1}}{2^{n+2}}$$ And so $\frac{c_k}{s_k} \le \frac{2^{n+2}}{k} \rightarrow 0$ as $k \rightarrow \infty$.

From the claim, this means $\frac{o_k}{s_k} \rightarrow \frac{1}{2}$ and so the sequence of fractions $\frac{o_ka + e_kb}{s_k}$ approaches $\frac{a+b}{2}$ as expected.

Geometric case: By the formula for sum of geometric series, we can show that $\frac{c_k}{s_k} = \frac{m^{k-1}}{(m^{k} - 1)/(m - 1)} \ge \frac{m-1}{m}$; hence it cannot go to 0. By the claim, the series of fractions diverges.

[More specifically I think it can be shown that $\frac{o_k}{s_k}$ oscillates between $\frac{1}{m+1}$ and slightly larger than $\frac{m}{m+1}$, using the facts that for even $k$, $e_k = mo_k$ and for odd $k$, $o_k = me_k + c_1$]