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

  • Thread starter Thread starter tmt1
  • Start date Start date
  • Tags Tags
    Infinity
tmt1
Messages
230
Reaction score
0
I have $$5^n - 3^n \le 2^n$$ (as n approaches infinity) but I'm not sure how to prove this to myself.
 
Mathematics news on Phys.org
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.
 

Thread 'What Exactly is Dirac’s Delta Function? - Insight'

Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
View full post »

Thread 'Fermat's Last Theorem'

Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
View full post »

Thread 'Useless continued fraction for 1'

I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
View full post »

Similar threads

A Limit of ##i^\frac{1}{n}## as ##n \to \infty##
Replies
14
Views
2K
MHB Can the Inequality of the Sum be Proven Using the Cube Root of -1?
Replies
4
Views
1K
B How to prove this for every Arithmetic Progression?
Replies
12
Views
1K
B How does the Riemann Hypothesis/Riemann Zeta function even work?
Replies
17
Views
841
A Proving Goldbach's conjecture by induction (or why it doesn't seem to work)
Replies
10
Views
2K
Prove that the limit as n->infinity of n^n/n! is infinity
Replies
7
Views
1K
B Sum of Series 1/n: Is it Infinity?
Replies
8
Views
2K
MHB Proving $\dfrac{(n-1)^{2n-2}}{(n-2)^{n-2}}<n^n$ for $n\ge 3$
Replies
1
Views
868
MHB Polynomial in n variables: Prove the identity
Replies
1
Views
1K
MHB Proving Inequality for All $n \ge 1$
Replies
3
Views
2K

Hot Threads

Recent Insights

Back
Top