Mean of two numbers by infinite sequences 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? 

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