Limit's of Sequence Help needed!

[djdavis2k]

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

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

b.) a_n= ((n!)^2)/(2n)!

c.) a_n=((n^2)*(2^n))/(n!)

d.) a_n= ((1/(3^5)),(1/(3^6)),(1/(3^7)),(1/(3^8)),....)

e.) a_n= sqrt(n^2+3*n)- n

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


for a.) if you divide the term by 'n' you result in a limit of 1... i'm not sure this is correct.. and i'm not sure what definition and theorem to use to support this conclusion

for b.) i believe if isolate the n! term you left with n!/2.. therefore it approaches infinity... so the limit does not exist.. i'm not sure what theorem or definition i can use to support my conclusion

for c.) if i expand 'n!' to n(n-1)! i get n^n/(n-1)!...... therefore I believe the limit does not exist because the number continous to grow larger as the denominator decrease... i'm not sure if my calculation for this is correct.. and what theorem and definition i can use to support this claim

for d.) i believe the expression can be simplifed to 1/(3^(5+n)) and assuming n increases .. this limit should approach zero... i'm not sure what theorem or definition to use to support this...


for e.) I believe the limit approaches -infinity so the limit does not exist as n increases... i'm not sure which test or definition to use here

for f.) i believe the limit does not exist because (-1)^n is a part of a harmonic series. so it will alternate between positive and negative values.. so a limit should not exist.. again I don't know which test or theorem or definiton to use to mathematically prove this

Please help me with these problems.. i believe i have to use the limit comparison test, alternating series, harmonic series, geometric series, or ratio test... to explain the conclusions but i'm not sure.. please help me with these problems

 

[PAULK]

Evaluate the limit (if it exists) of each of the following sequences. Indicate the results (definition,theorems,etc.) you use to support your conclusion.
a.) a_n= {(n-3)/(n)}^n

Your a: looks like this;

a_n = (1 - 3/n)^n

Compare this with

b_n = (1 + 1/k)^k which is known --> e.

Try the substitution 1/k = -3/n and do some algebra. That should do it.

 

[djdavis2k]

Evaluate the limit (if it exists) of each of the following sequences. Indicate the results (definition,theorems,etc.) you use to support your conclusion.
a.) a_n= {(n-3)/(n)}^n

Your a: looks like this;

a_n = (1 - 3/n)^n

Compare this with

b_n = (1 + 1/k)^k which is known --> e.

Try the substitution 1/k = -3/n and do some algebra. That should do it.

therefore if that is equivalent to 'e'... doesn't the limit approach infinity since e^inf. is infinity.... how about b to f.. is my approach correct... i need help there

 

[soroban]

Hello, djdavis2k!

I believe (b) converges . . .

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

\(\displaystyle \text{We have: }\:a_n \;=\;\frac{(n!)(n!)}{(2n)!} \;=\; \frac{n!}{n!}\cdot\frac{n!}{(n+1)(n+2)(n+3)\hdots 2n}\)

. . \(\displaystyle = \;\frac{1}{n+1}\cdot\frac{2}{n+2}\cdot\frac{3}{n+3} \hdots\frac{n-1}{2n-1}\cdot\frac{n}{2n}\)

\(\displaystyle \text{While the last half of the fractions approach }\frac{1}{2}\text{, the first half approaches }0.\)

\(\displaystyle \text{Therefore: }\;\lim_{n\to\infty}a_n \;=\;0\)

 

[PAULK]

Evaluate the limit (if it exists) of each of the following sequences. Indicate the results (definition,theorems,etc.) you use to support your conclusion.
a.) a_n= {(n-3)/(n)}^n

Your a: looks like this;

a_n = (1 - 3/n)^n

Compare this with

b_n = (1 + 1/k)^k which is known --> e.

Try the substitution 1/k = -3/n and do some algebra. That should do it.

therefore if that is equivalent to 'e'... doesn't the limit approach infinity since e^inf. is infinity.... how about b to f.. is my approach correct... i need help there

No, it does not. Try this: 1/k = -3/n, so k = - n/3, n = -3k. then

a_n = (1 - 3/n)^n = (1 + 1/k)^(-3k) = [(1 + 1/k)^3k] ^-3 = ???

....................................
OK, here is another:
e.) a_n= sqrt(n^2+3*n)- n can be done by your old friend: rationalizing a denominator: [OK, OK, it's a numerator, but who's counting?]
........sqrt(n^2+3*n)- n sqrt(n^2+3*n) + n
a_n = --------------------- ---------------------- =
............................ sqrt(n^2+3*n) + n

n^2 + 3n - n^2
---------------------- =
sqrt(n^2+3*n) + n

3n
---------------------- =
sqrt(n^2+3*n) + n


Now use the 'complex fraction trick' -- divide all terms by n:

3
--------------------- = ... you can do this now.
sqrt(1+3/n) + 1

 

[djdavis2k]

thanks for your help

for c.) is this the correct approach .... if i expand 'n!' to n(n-1)! i get n^n/(n-1)!...... therefore I believe the limit does not exist because the number continous to grow larger as the denominator decrease... i'm not sure if my calculation for this is correct.. and what theorem and definition i can use to support this claim

for d.) is this the correct approach....i believe the expression can be simplifed to 1/(3^(5+n)) and assuming n increases .. this limit should approach zero... i'm not sure what theorem or definition to use to support this...


for. f. it's a harmonic series so it should alternate between positive and negative values so the limit should not exist right because of the (-1)^n factor....


please help me with these problems