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

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SUMMARY

The inequality $$5^n - 3^n \le 2^n$$ is proven to be false as n approaches infinity. The discussion clarifies that for sufficiently large n, specifically for n ≥ 1, the expression $$5^n - 3^n$$ exceeds $$2^n$$. This conclusion is supported by the binomial expansion of $$5^n$$, which shows that it grows faster than both $$2^n$$ and $$3^n$$, confirming that the inequality does not hold true.

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I have $$5^n - 3^n \le 2^n$$ (as n approaches infinity) but I'm not sure how to prove this to myself.
 
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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 $5^n-3^n > 2^n$
 
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 $x\ge 1$, $5^x- 3^x\ge 2^x$.
 
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.
 

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