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

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The inequality $5^n - 3^n \le 2^n$ is incorrect as n approaches infinity. Using binomial expansion, it can be shown that $5^n$ is greater than $2^n + 3^n$, leading to the conclusion that $5^n - 3^n > 2^n$. The discussion clarifies that the original poster meant the inequality holds for sufficiently large n, but it remains false for all n greater than or equal to 1. Therefore, the assertion does not hold true as n increases. The conclusion is that the inequality does not hold for large values of n.
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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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