Calculus finding a limit help please!?

Question

I am solving the series from 1 to infinity of

1 * 3 * 5 .....(2n-1)
-------------------------
1*4*7.......(3n-2)


using the ratio test of | An + 1 |
--------------
| An |

and I am getting:

(2n+1) (3n-2)
---------- X ---------
(3n+1) (2n-1)


to figure out if the series converges don't i just take the limit of what I wrote above? Is the limit 1? Show work please.

Thank you!

Answer 1

if the ratio of terms converged to one, then your series would ultimately be close to
sum of earlier terms + a[N]*(1+1+1+.....) which would not converge.

But it doesn't:
A[n+1]/A[n] = (2n+1) / (3n+1) = ( (2/3)(3n+1) + (1/3) ) /(3n+1)
= (2/3) + 1/(3(3n+1)) -->2/3

now by taking a large enough number of initial terms your series will be close to sum of earlier terms + a[N]*(1+(2/3)+(2/3)^2+...) =
sum earlier terms + a[N]*3
and will converge

Answer 2

Wow , this is considered a very hard question in calculus as far as my knowledge goes ( I'm in High school )
( pardon my English , I'm not really familiar with mathematical terms in English )
I don't know the exact solving method but :
1) THIS LIMIT DOES HAVE AN ANSWER !
2) The answer is NOT 1 cause the result keeps getting lower as more "n" gets .
3) The answer is NOT 2/3 because as "n" gets bigger the answer keeps getting much lower than 2/3 ( try n=3 and even then it's less than 2/3 )
I saw an answer said 2/3 which he/she mistakenly solved this problem rather than yours :
(2n+1)
Lim -------- when n ----> infinity
(3n-2)
which is a common mistake !

4) Writing down the problem like :
(2n+1)(3n-2)
------------------
(3n+1)(2n-1)

only indicates that this figure keeps getting less so it can not be 1 ? ( Why the hell are you getting a limit from a ratio test ??!! )

==> My conclusion is that since the figure is getting less and less ( Each time your multiplying it by a number less than 1 so it keeps getting less that the initial figure "1" ) and if you write down some of the sentences :
( 1 , 3/4 , 15/28 , 105/280 ...)
it indicates the limit is headed towards zero .
My guess is that ZERO is the answer .

Regards
Soheil

P.S. Email me if you had another problem .

Answer 3

Just focus on calculating limit.

lim (2n-1)/(3n-2) as n--> infinity
= lim (2-(1/n)) / (3-(2/n)) as n--> infinity
= (2 - 0)/(3- 0) =2/3

since (2/3) < 1 therefore the series converges.
ok.

Answer 4

| An + 1 |
-------------- =
| An |

=

(2n+1)
---------- -> (n-> oo)
(3n+1)

2/3 < 1 => the series converge (D'Alambers criterion)

Answer 5

What you wrote is OK.
Numerator and denominator are polynomes of 2nd degree
To find the limit n=> infinity you need to look only at the 2nd degree coefficients which are 6 both.So the limit is 1 and the
sole limit wont help

Answer 6

This looks nasty, yet its fairly no longer frustrating. Plug in 0 into x. We get sin(2cos0)/5sec0. cos0=one million, so 2cos0=2. 5sec0=5/cos0. Cos0=one million, so we get 5. Dividing those 2 factors, we get sin2/5. To do those problems continuously first plug in x= in spite of type it procedures. This time there grew to become no longer something nasty in touch, basically subject-loose trig.