Limits: a_n = { [(n-3)/(n)]^3 }, a_n= { [((n^2)(2^n))/(n!)]}

[djdavis2k]

Evaluate the limit (if it exist) of each of the following sequences. Indicate the results (definitions,theorems,etc.) you use to support your conclusion.

a.) a_n = { [(n-3)/(n)]^3 }

b.) a_n= { [((n^2)(2^n))/(n!)]}

c.) a_n= {1/3^5, -1/3^6,1/3^7,-1/3^8,....}

d.) a_n= {((-1)^n)*((2n^3)/(n^3+1))}

my work shown:

for a.) if you divide the numerator and the denominator of the expression within the brackets of the exponent of 'n' by n you get: a_n= (1- (3/n))^n

if you use a comparison test with b_n= (1+(1/k))^k and let 1/k= -3/n and k=-n/3 and n=-3k

therefore a_n becomes: (1+1/k)^(-3k)

if i take the natural logarithm of both sides i get ln a_n= = -3k (ln (1+1/k)

since the lim of n--> infinity of 1/k equals 0.. .therefore the expression (ln (1+1/k)= ln 1= 0

however, as k approaches infinity, -3k approaches -negative inf.

so the expression results in -infinity*zero= 0 so e^0= 1

so the limit approaches 1... i believe i made a few mistakes here... please help me out


for b.) a_n= (n^2)(2^n)/(n!) can be simplified to (n*2^n)/(n-1)! since n! can also be expressed as n(n-1)!

if the sequence is analyzed the denominator becomes much larger than the num. as n increases from n equals 1

if lim of n approaches infinity of (n)(2^n)/(n-1)! approaches zero.. but how do i explain this mathematically

can i say a_n can be compared to b_n= 1/n and since a_n is less than b_n it can be used to explain the sequence of a_n.. therefore it is equal to zero
i need some help here.. what did i do wrong


for c.) the expression can be simplified to a_n= (-1)^(n+1)) (1/3^(4+n)) assuming n=1 ... since this is symbolic of a harmonic series

can i say a_(n+1)> a(n), a(n+2)>a(n+1).. therefore the sequence is an alternating sequence.. so the sequence does not approach a value as n approaches infinity.. so the limit does not exist... how do i write this mathematically and algebracially... need a little bit of help here

for d.) a_n= ((-1)^n))*( (2n^3)/(n^3+1)) it can be simplified by dividing the number of the expression by the highest power factor which is "n^3"
which results in 2/((1+(1/n^3)))* (-1)^n
since, the limit of of n approaches of 1/n^3 approaches 0 since (1/n^3)<(1/n^2(<(1/n)

so the a_n expression becomes a_n= (-1)^n*(2)= {-2,2,-2,2,-2}
so the sequences is alternating between -2 and 2 .. so the sequence doesn't reach a limit as n approaches infinity... so the limit does not exist..
how do i express this mathematically and algebrically as well..

 

[soroban]

Re: Limit's Help.... work shown please help!!!

Hello, djdavis2k!

Thank you for showing your work!

Evaluate the limit (if it exist) of each of the following sequences.

\(\displaystyle a)\;\;a_n \:=\:\left(\frac{n-3}{n}\right)^3\)

Your work is admirable, but you misread the problem . . . the exponent is "3".

\(\displaystyle \text{So we have: }\:a_n \:=\:\left(\frac{n-3}{n}\right)^3 \;=\;\left(\frac{n}{n} - \frac{3}{n}\right)^3 \;=\; \left(1 - \frac{3}{n}\right)^3\)

\(\displaystyle \text{Then: }\:\lim_{n\to\infty}\left(1 - \frac{3}{n}\right)^3 \;=\;(1-0)^3 \;=\;1\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

\(\displaystyle \text{If the exponent was indeed an }n\text{, then: }\;\lim_{n\to\infty}\left(1 - \frac{3}{n}\right)^n \;=\;e^{-3}\)

\(\displaystyle b)\;\; a_n \:=\:\frac{n^2\cdot2^n}{n!}\)

\(\displaystyle \text{Are you allowed the Ratio Test? }\;R \;=\;\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\)


\(\displaystyle \frac{a_{n+1}}{a_n} \;=\;\frac{\dfrac{(n+1)^2\cdot2^{n+1}}{(n+1)!}} {\dfrac{n^2\cdot2^n}{n!}} \;=\;\frac{(n+1)^2\cdot2^{n+1}}{(n+1)!}\cdot\frac{n!}{n^2\cdot2^n} \;=\;\frac{(n+1)^2}{n^2}\cdot\frac{2^{n+1}}{2^n} \cdot\frac{n!}{(n+1)!}\)

. . . . \(\displaystyle = \;\left(\frac{n+1}{n}\right)^2\cdot 2\cdot\frac{1}{n+1} \;=\;\left(1 + \frac{1}{n}\right)^2\cdot\frac{2}{n+1}\)


\(\displaystyle \text{Hence: }\:\lim_{n\to\infty}\left[\left(1+\frac{1}{n}\right)^2\cdot\frac{2}{n+1}\right] \;=\;1^2\cdot0 \;=\;0\)

\(\displaystyle \text{Therefore, the sequence converges to zero.}\)

\(\displaystyle c)\;\;a_n \:=\:\left\{ \frac{1}{3^5},\:\text{-}\frac{1}{3^6},\:\frac{1}{3^7},\:\text{-}\frac{1}{3^8}\:\hdots \right\}\)

\(\displaystyle \text{You are correct: }\:a_n \:=\:\frac{(-1)^{n+1}}{3^{n+4}}\)

\(\displaystyle \text{But it is a }geometric\text{ sequence, and it }converges\text{ to zero.}\)

\(\displaystyle d)\;\; a_n\:=\-1)^n\,\frac{2n^3}{n^3+1}\)

\(\displaystyle \text{Again, you are correct: }\:a_n \:=\-1)^n\,\frac{2}{1 + \frac{1}{n^3}}\)

\(\displaystyle \text{And its limit is: }\:\lim_{n\to\infty} a_n \:=\:\pm2\)

. . Therefore, the sequence diverges.

 

[johnq2k7]

Re: Limit's Help.... work shown please help!!!

Hello, djdavis2k!

Thank you for showing your work!

Evaluate the limit (if it exist) of each of the following sequences.

\(\displaystyle a)\;\;a_n \:=\:\left(\frac{n-3}{n}\right)^3\)

Your work is admirable, but you misread the problem . . . the exponent is "3".

\(\displaystyle \text{So we have: }\:a_n \:=\:\left(\frac{n-3}{n}\right)^3 \;=\;\left(\frac{n}{n} - \frac{3}{n}\right)^3 \;=\; \left(1 - \frac{3}{n}\right)^3\)

\(\displaystyle \text{Then: }\:\lim_{n\to\infty}\left(1 - \frac{3}{n}\right)^3 \;=\;(1-0)^3 \;=\;1\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

\(\displaystyle \text{If the exponent was indeed an }n\text{, then: }\;\lim_{n\to\infty}\left(1 - \frac{3}{n}\right)^n \;=\;e^{-3}\)



[quote:1zqrkaft]\(\displaystyle b)\;\; a_n \:=\:\frac{n^2\cdot2^n}{n!}\)

\(\displaystyle \text{Are you allowed the Ratio Test? }\;R \;=\;\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\)


\(\displaystyle \frac{a_{n+1}}{a_n} \;=\;\frac{\dfrac{(n+1)^2\cdot2^{n+1}}{(n+1)!}} {\dfrac{n^2\cdot2^n}{n!}} \;=\;\frac{(n+1)^2\cdot2^{n+1}}{(n+1)!}\cdot\frac{n!}{n^2\cdot2^n} \;=\;\frac{(n+1)^2}{n^2}\cdot\frac{2^{n+1}}{2^n} \cdot\frac{n!}{(n+1)!}\)

. . . . \(\displaystyle = \;\left(\frac{n+1}{n}\right)^2\cdot 2\cdot\frac{1}{n+1} \;=\;\left(1 + \frac{1}{n}\right)^2\cdot\frac{2}{n+1}\)


\(\displaystyle \text{Hence: }\:\lim_{n\to\infty}\left[\left(1+\frac{1}{n}\right)^2\cdot\frac{2}{n+1}\right] \;=\;1^2\cdot0 \;=\;0\)

\(\displaystyle \text{Therefore, the sequence converges to zero.}\)

\(\displaystyle c)\;\;a_n \:=\:\left\{ \frac{1}{3^5},\:\text{-}\frac{1}{3^6},\:\frac{1}{3^7},\:\text{-}\frac{1}{3^8}\:\hdots \right\}\)

\(\displaystyle \text{You are correct: }\:a_n \:=\:\frac{(-1)^{n+1}}{3^{n+4}}\)

\(\displaystyle \text{But it is a }geometric\text{ sequence, and it }converges\text{ to zero.}\)

\(\displaystyle d)\;\; a_n\:=\-1)^n\,\frac{2n^3}{n^3+1}\)

\(\displaystyle \text{Again, you are correct: }\:a_n \:=\-1)^n\,\frac{2}{1 + \frac{1}{n^3}}\)

\(\displaystyle \text{And its limit is: }\:\lim_{n\to\infty} a_n \:=\:\pm2\)

. . Therefore, the sequence diverges.[/quote:1zqrkaft]

thanks for your help... are you sure a.) is correct because for the exponent of 'n'
for a.) shouldn't the value of the lim of n--> inf of {(1-3/n)}^n
be 0 or e^0 how did u get e^-3.... since 3/n as n approaches inf. is 0... and the ln 1 is 0
so if the natural logarithm is taken of a_n it equals n (ln (1-3/n) which is then equal to n*(0) so the e^(n*0) is equal to 1.. or the value is zero i'm not sure... please help me understand how u got e^-3

 

[galactus]

Re: Limit's Help.... work shown please help!!!

Regarding Soroban's \(\displaystyle e^{-3}\), you should know the famous limit \(\displaystyle \lim_{n\to \infty}\left(1+\frac{1}{x}\right)^{n}=e\)

You don't have to worry about proving this each time. Just use it. See now why it's \(\displaystyle e^{-3}\)?.

Also, \(\displaystyle \lim_{n\to 0}(1+n)^{\frac{1}{n}}=e\)

These will prove handy.