Using the definition of convergence, how do you prove that the sequence # lim (3n+1)/(2n+5)=3/2# converges?

Question

Using the definition of convergence, how do you prove that the sequence # lim (3n+1)/(2n+5)=3/2# converges?

1 Answer

Related questions

What is the sum of the series #1+ln2+(((ln2)^2)/(2!))+...+(((ln2)^n)/(n!))+...#?
How do I write #(5/(1*2))+(5/(2*3))+(5/(3*4))+...+(5/n(n+1))+...#in summation notation, and how...
How do you show whether the improper integral #int ln(x)/x^3 dx# converges or diverges from 1 to...
How do you show whether the improper integral #int e^x/ (e^2x+3)dx# converges or diverges from 0...
How do you show whether the improper integral #int (79 x^2/(9 + x^6)) dx# converges or diverges...
How do you show whether the improper integral #int 1/ (1+x^2) dx# converges or diverges from...
How do you prove that the integral of ln(sin(x)) on the interval [0, pi/2] is convergent?
How do you show whether the improper integral #int (x^2)(e^(-x^3)) dx# converges or diverges...
Using the definition of convergence, how do you prove that the sequence #{5+(1/n)}# converges...
Using the definition of convergence, how do you prove that the sequence #{2^ -n}# converges from...

Impact of this question

4911 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-06T15:53:57

I assume meant limit at infinity, that is, to show that #lim_(n->infty) (3n+1)/(2n+5) = 3/2#. Note that #lim_(n->infty)1/n = 0# and similarly #lim_(n->infty)k/n = 0# for any real, positive #k#.

Explanation:

Also note that by dividing by #n# on both numerator and denominator, #(3n+1)/(2n+5) = (3+1/n)/(2+5/n)#

This means that #lim_(n->infty) (3n+1)/(2n+5) = lim_(n->infty) (3+1/n)/(2+5/n) = (3+0)/(2+0) = 3/2#

Science

Math

Humanities

... and beyond

Using the definition of convergence, how do you prove that the sequence # lim (3n+1)/(2n+5)=3/2# converges?

1 Answer

Explanation:

Related questions

Impact of this question