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

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Discussion Overview

The discussion revolves around the inequality $$5^n - 3^n \le 2^n$$ as n approaches infinity. Participants explore the validity of this inequality, questioning its truth and the implications of the limit as n increases.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the inequality $$5^n - 3^n \le 2^n$$ holds as n approaches infinity, but express uncertainty about how to prove it.
  • Others argue against the inequality, suggesting that using the binomial expansion shows that $$5^n > 2^n + 3^n$$, leading to the conclusion that $$5^n - 3^n > 2^n$$.
  • A participant questions the meaning of "as n goes to infinity," suggesting that it typically refers to the limit, and asks if the intent was to imply the inequality holds for sufficiently large n.
  • Another participant reinforces the argument against the inequality, stating that for values of x less than or equal to 1, the inequality holds, but for x greater than or equal to 1, it does not.
  • One participant clarifies that they indeed meant "sufficiently large n" in their original statement.

Areas of Agreement / Disagreement

Participants do not reach consensus on the validity of the inequality. There are competing views, with some asserting it is true and others providing counterarguments that it is not.

Contextual Notes

The discussion highlights the importance of defining the conditions under which the inequality is considered, particularly regarding the values of n and the implications of limits.

tmt1
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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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