How do you use the ratio test to test the convergence of the series #∑ ((4n+3)^n) / ((n+7)^(2n))# from n=1 to infinity?

Answer 1

You use the Cauchy test

#sum_(n=1)^oo a_n = sum_(n=1)^(oo)((4n+3)^n) / ((n+7)^(2n))#
if #lim_(n->oo)a_n^(1/n) < 1# then the series converge
if #> 1# then it diverge
if #= 1# you can't conclude
#(((4n+3)^n) / ((n+7)^(2n)))^(1/n) = (4n+3)/(n+7)^2 = (4n+3)/(n^2+14n+49) = (4+3/n)/(n(1+14/n+49/n^2)#

take the limit

#lim_(n->oo) (4+3/n)/(n(1+14/n+49/n^2)) = 0#

so it converge

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

According to the root test, the series converges.

I know that you have asked for the ratio test to test the convergence.

However, in this case, I would strongly recommend to do the root test instead.

You can transform your series as follows:

# sum_(n=1)^(oo) (4n +3)^n/(n+7)^(2n) = sum_(n=1)^(oo) (4n+3)^n/((n+7)^2)^n = sum_(n=1)^(oo) ((4n+3)/((n+7)^2))^n#
The root test states that for a series #sum_(n=1)^(oo) a_n#,
As you can transform your #a_n# as an expression taken to the power of #n#, the root test is really well suited.
#lim_(n->oo) root(n)(abs(a_n)) = lim_(n->oo) root(n)(abs( ((4n+3)/((n+7)^2))^n ))#
You can omit the absolute value since #((4n+3)/((n+7)^2))^n > 0# is true for any #n > 0#. Thus, your calculation simplifies to:
#lim_(n->oo) root(n)(abs(a_n)) = lim_(n->oo) (4n+3)/((n+7)^2)#
# = lim_(n->oo) (4n+3)/(n^2 + 14n + 49)#
... the highest power of #n# in your fraction is #n^2#. Factor #n^2# in the numerator and denominator...
# = lim_(n->oo) (n^2 (4/n +3/n^2 ))/(n^2( 1 + 14/n + 49/n^2))#
# = lim_(n->oo) (cancel(n^2) (4/n +3/n^2 ))/(cancel(n^2)( 1 + 14/n + 49/n^2))#
# = lim_(n->oo) ( 4/n +3/n^2 )/( 1 + 14/n + 49/n^2)#

... take the limit...

# = ( 0 + 0 )/( 1 + 0 + 0)#
# = 0#
As #lim_(n->oo) root(n)(abs(a_n)) = 0 < 1#, according to the root test, the series converges.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3
To use the ratio test to test the convergence of the series \(\sum \frac{(4n+3)^n}{(n+7)^{2n}}\) from \(n=1\) to \(\infty\), follow these steps: 1. Compute the ratio of consecutive terms: \[r = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\] where \(a_n\) represents the general term of the series. 2. Plug in the general term: \[a_n = \frac{(4n+3)^n}{(n+7)^{2n}}\] 3. Compute \(a_{n+1}\) by replacing \(n\) with \(n+1\) in the general term expression. 4. Calculate the ratio \(r\) by taking the limit as \(n\) approaches infinity. 5. Determine the convergence of the series based on the value of \(r\): - If \(r < 1\), the series converges absolutely. - If \(r > 1\), the series diverges. - If \(r = 1\), the test is inconclusive, and another test may be needed. 6. If the ratio test gives a result of \(r < 1\), you can conclude that the series converges absolutely. If \(r > 1\), you can conclude that the series diverges. If \(r = 1\), the ratio test is inconclusive, and other convergence tests or methods should be used to determine the convergence or divergence of the series.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7