0

I've simply said that adding $1$ to the numerator and denominator when the denominator is $5^n$ (greater than $2^n$), the "$+1$" accounts for a larger portion of the numerator than the denominator, so $\frac{2^n+1}{5^n+1}$ better be greater than$\frac{2^n}{5^n}$ for all $n\gt0$.

I'm just not satisfied with this explanation because it seems sloppy and not rigorous... Does someone have a better proof?

Cotton Headed Ninnymuggins
  • 840
  • 1
    Use the fact that if $bd>0$, then $\frac{a}b>\frac{c}d$ if and only if $ad>bc$: you can convert the first inequality to the second by multiplying both sides by $bd$, since multiplying by a positive number does not change the direction of the inequality. – Brian M. Scott Apr 06 '21 at 05:42
  • 1
    What you say is correct and you can prove that $(a+1)/(b+1)>a/b$ is equivalent to $a<b$ (assuming $a, b$ are positive). – Paramanand Singh Apr 06 '21 at 06:42
  • Also: https://math.stackexchange.com/q/543682/42969, https://math.stackexchange.com/q/734128/42969 – Martin R Apr 06 '21 at 08:05

3 Answers3

3

Just show that $LHS - RHS>0$.

$$\frac{2^n+1}{5^n+1}-\frac{2^n}{5^n} = \frac{10^n+5^n-10^n-2^n}{5^n(5^n+1)} = \frac{5^n-2^n}{5^n(5^n+1)} >0$$

(As $5^n>2^n \forall n>0$ and the denominator is always positive.)

Or $\dfrac{2^n+1}{5^n+1} >\dfrac{2^n}{5^n}$

19aksh
  • 12,768
3

$$\frac{2^n+1}{5^n+1} >\frac{2^n}{5^n}$$ $$ \Leftrightarrow (2^n+1)5^n > (5^n+1)2^n$$ $$ \Leftrightarrow 2^n 5^n + 5^n > 2^n 5^n + 2^n $$ $$ \Leftrightarrow 5^n > 2^n $$ which is true for $n>0$

Blitzer
  • 2,018
0

Think batting averages. $\frac{2^n+1}{5^n+1}$ is some sort of weighted average of $\frac{2^n}{5^n}$ and $\frac 11 = 1$, so it must be greater than $\frac{2^n}{5^n}$.

Robert Shore
  • 23,332
  • 1
    I'm not following this, what if the batting average was $\frac{5^n}{2^n}$? now that would not be greater than $\frac{5^n+1}{2^n+1}$ just because $\frac{1}{1}=1$ – Cotton Headed Ninnymuggins Apr 06 '21 at 06:35
  • And it's true that $\frac{5^n}{2^n} \gt \frac {5^n+1}{2^n+1}$. Now you're taking a weighted average of a number greater than $1$ with $1$. In general, it's pretty easy to prove that if $\frac ab \lt \frac cd$, then $\frac ab \lt \frac{a+c}{b+d} \lt \frac cd$. – Robert Shore Apr 06 '21 at 09:01
  • I also meant to say it the other way around and that the “$+1$’s” don’t make $\frac{5^n}{2^n}$ any bigger, but actually makes it smaller – Cotton Headed Ninnymuggins Apr 06 '21 at 20:36
  • Again, that's because you're taking a weighted average of $1$ with a number that's greater than $1$. You'll end up with something in between them. – Robert Shore Apr 07 '21 at 00:50