Question

limit n tend to( n/2^n)the sequences is converge or diverg

Answer 1

as n tends to infinity?

Answer 2

use l'Hospiatals rule, do you know it?

Answer 3

yes n tends to infinity

Answer 4

I recommend using l'Hospital if you know it

Answer 5

yes i know it hotial rule but procedure dont know . i mean step

Answer 6

when you have a form like\[\pm\frac\infty\infty\]or\[\frac00\]you can take the derivative of the top and bottom, then try taking the limit again
this is called l'Hospital's Rule

what is the derivative of the numerator?

Answer 7

top derivative is zero and bottom../?

Answer 8

this is l'Hospital in action:\[\lim_{n\to\infty}\{a_n\}=\lim_{n\to\infty}{n\over2^n}=\lim_{n\to\infty}{\frac d{dn}(n)\over\frac d{dn}(2^n)}\]the derivative of n with respect to n is not zero
please try again

Answer 9

what is the derivative of 2^n

Answer 10

\[\frac d{dx}(a^x)=a^x\ln a\]

Answer 11

but you still haven't told me what the derivative of n is

Answer 12

derivative of n is 1..?

Answer 13

yes :)

Answer 14

and what is the derivative of 2^n ?

Answer 15

sorry sir i can't understand.. can u explain ..:(

Answer 16

\[\frac d{dx}(a^x)=a^x\ln a\]you do not understand this formula?

Answer 17

no

Answer 18

\[\frac d{dx}(a^x)=a^x\ln a\]examples:\[\frac d{dx}(5^x)=5^x\ln 5\]\[\frac d{dx}(132^x)=132^x\ln (132)\]\[\frac d{dx}(\pi^x)=\pi^x\ln\pi\]what is \(a\) in your case?

Answer 19

2

Answer 20

right, so applying the formula\[\frac d{dx}(a^x)=a^x\ln a\]to\[\frac d{dx}(2^x)\]you get what?

Answer 21

2^xln2

Answer 22

exactly :)

Answer 23

but in our case 2^nln(n) and the value of is infinity..

Answer 24

why do you say ln(n) ?
it will be ln2

Answer 25

Remember how l'Hospital works: derivative of top and bottom separately when we have \(\pm\frac\infty\infty\) or \(\frac00\)
so the derivative of the top is 1
the derivative of the bottom is (2^x)ln(2)
so we have\[\lim_{n\to\infty}\{a_n\}=\lim_{n\to\infty}{n\over2^n}=\lim_{n\to\infty}{\frac d{dn}(n)\over\frac d{dn}(2^n)}=\lim_{n\to\infty}\frac1{2^n\ln 2}\]and that limit you can take, and it is...?

Answer 26

2^n and n tends to infinity the 2^n i also infinity

Answer 27

ln2 is 0.69

Answer 28

ln(2) is a constant, so it doesn't tend to anything
2^n tends to infinity, so the whole fraction tends to...?

Answer 29

infinity. bottom value is infinity ans1/infinity is zero. so our sequence is converg

Answer 30

yep :)

Answer 31

than u so much what a explain.. thanks jazak allah

Answer 32

you are welcome!
please ask if you need more clarification

Answer 33

nsin(pi/n)

Answer 34

once again, I ask you to please post every question separately
thank you

Answer 35

it is also considered polite to click the button that says "best answer" if you like that answer
not that I care, but it is how the site works

Answer 36

thanks again i am very thankful to u