Proving $5^n - 3^n \le 2^n$ as n approaches Infinity

tmt1 · Jul 21, 2016

I have $$5^n - 3^n \le 2^n$$ (as n approaches infinity) but I'm not sure how to prove this to myself.

kaliprasad · Jul 22, 2016

tmt said:

I have $$5^n - 3^n \le 2^n$$ (as n approaches infinity) but I'm not sure how to prove this to myself.

Not true. because
$5^n = (2+3)^n = 2^n + 3^n$ + some positive terms using binomial expansion
hence
$5^n > 2^n+3^n$
or $5^n-3^n > 2^n$

HallsofIvy · Jul 22, 2016

What, exactly do you mean by an inequality in n, "as n goes to infinity"? Normally, "as n goes to infinity" means "in the limit as n goes to infinity" but that cannot be what is meant here because your inequality depends on specific n. Do you mean "the inequality is true for sufficiently large n"? In any case, as kalisprasad said, this is simply not true. In fact, for $x\le 1$, $5^x- 3^x\le 2^x$ but for all $x\ge 1$, $5^x- 3^x\ge 2^x$.

tmt1 · Jul 22, 2016

HallsofIvy said:

What, exactly do you mean by an inequality in n, "as n goes to infinity"? Normally, "as n goes to infinity" means "in the limit as n goes to infinity" but that cannot be what is meant here because your inequality depends on specific n. Do you mean "the inequality is true for sufficiently large n"? In any case, as kalisprasad said, this is simply not true. In fact, for $x\le 1$, $5^x- 3^x\le 2^x$ but for all $x\ge 1$, $5^x- 3^x\ge 2^x$.

yes, I mean sufficiently large n.

Proving $5^n - 3^n \le 2^n$ as n approaches Infinity

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