Exam 3&4 Cal II
what is the structure of a geometric series?
(fraction)^n
as k approaches infinity for lnk/k, the limit is equal to ___
0
because r has to be less than 1 for a geometric series to converge, a series of cos^2x does not converge when the values are equal one, at the values __ and __
0 and pi
what is the p value of the series 5/k^k (lnk+2)^k?
0 because by the root test, ak^1/k = 5^1/k (lnk/k + 2/k) (after division of k) and the limit of ak as it approaches 0 equals 1 * (0+0)
the series 1/n *ln(4/6n+2) will always approach __ because the addition of 1/n causes
0, the entire sequence to approach 0
the series 2 sin (npi) always ___ because the numerator will always be __
converges (to zero), 0
the series (2sinn^2)^2/(5n^2 +4) ___ by the ___ test because (sin(n))^2 is less than or equal to one therefore we can compare the series to the function 1/n^2 that is greater than and converges by the p-series test
converges, comparison
the series 2ln(4n)/n^2 ___ by the ___ test
converges, integral
if the summation of | a sub n | is convergent then a sub n would be absolutely convergent
definition of absolute convegence
the series 5n/2n^2+6 ___ because...
diverge, it greater than the series 1/n that always diverges, meaning that it must always diverge
the summation from n=1 to infinity of the series 1/n is an example of ____
divergence
the series 1+m^4/sqrt(3+m^8) diverges by the ____ test because as m approaches infinity, the function is equal to __
divergence, 1
determine if the sequence converges when an= 3 + (-4/3pi)^n
since (-4/3pi)^n is equal to zero as n approaches infinity (4<3pi) then the sequence converges and has a limit of 3+0=3
the p value of the series of (5n+3/4n)^2n is equal to ____ because...
25/16, after dividing by n from top and bottom and raising the series to the power of 1/n (multiplying 1/n * 2n) take the limit of infinity of the series when it is squared, make sure to square the value at the end
what is the limit of the series n=1 to infinity of 3^nn!/n^n
3/e, (n+1)^n+1/n^n is simplified to (1+1/n)^n * n+1 (goes on the bottom)
what is the p value for the series 2^2n (n-4/n)^n^2?
4e^-4 steps: 1. divide by n from top and bottom 2. raise the function to the nth power 3. take the limit of the simplification, should be a value times (1-#/n)^n, meaning the limit should equal e 4. multiply the number outside by the value of e
which of the following series is convergent? 4/1+sqrt(n), (-1)^2n 1/4+sqrt(n), (-1)^n-1 5+sqrt(n)/1+sqrt(n), (-1)^n-1/5+sqrt(n)
4th series, the series 1st series is divergent by the comparison test to 1/p^1/2 (p<1), the 2nd series is divergent by the comparison test to 1/p^1/2 (because the negative gets canceled out by the exponent multiplied by 2, the 3rd series does not converge by the divergence test, and the 4th series converges because bn+1 is less than bn (the terms are dec) and as the limit approaches infinity the limit is zero
find the smallest number of terms of the series (-1)^k k/4^k needed to add in order to estimate with an error that is less than 7/4^7
6 terms, because the values dec as n inc (bn+1 < bn) then the value closest and less than 7/4^7 would have to be 6 terms
the p value of the series of 3n+4/n is ___ because
6/5, the limit of an^1/n is equal to 1 times the fraction
the p value for the series of the summation from n=1 to infinity of (7tan^-1(n)/3)^n is equal to ___ because...
7pi/6, an^1/n = 7tan^-1n/3 (without the n exponent) and the limit as n approaches infinity of tan^-1n = tan^1(infinity) = pi/2 (tangent is sin over cos to the value of pi/2 works because it is 1/0 being undefined)
if lim as n approaches infinity of an=8 then the limit of an-9 is approaches __
8 because changing one value is not large enough for the limit to be something other than infinite
what is the ratio test?
Use the ratio test if have n! do L=lim l an+1 / anl n→∞ If lim < 1, absolutely converges If lim > 1, diverges If lim = 1, test fails
does the alternating series test prove convergence?
YES
determine if the series n=0 to infinity of (2/3)^n/2
a=1 r= sqrt(2)/sqrt(3) meaning the series converges with a sum of a/1-r = sqrt(3)/sqrt(3)-sqrt(2)
3-9/2-27/4-81/8+... is
a=3, r=-3/2, divergent because the value of r will be greater than 1
how to find the sum of the finite series 6+(6*4)/9+(6*4^2)/9^2...+(6*4^8)/9^8
a=6, r=4/9, a(1-r^9/1-r)- the formula used to find the next value
the series (-1)^n-1 cos^2(n)/2^n is ___ because...
absolutely convergent, since cos^2(n) can only approach -1 to 1, the value of 1/2^n is most of the time larger than 1/2^n, and the geometric series always converges because r is less than 1
the series (-1)^n sin(5n)/2^n ___ because...
absolutely converges, the value of sin outputs neg and pos values, meaning it is sometimes less than 1/2^n so it must converge by the geometric test and the comparison test
the series of the summation of n=1 to infinity of (-7)^n/n! is ____ convergent because...
absolutely, the absolute value of the series is equal to 1 by the ratio test
determine whether the sequence an converges or diverges when an= (-1)^n(3n/5n+1), and if it does find its limit
after hospital, the limit approaches 3/5 meaning values of an oscillate between values ever closer to +-3/5, meaning the sequence diverges
find the sum whenever the series (cos^2x)^k
common ratio r= cos^2x because the series starts at k =1, so r/1-r, meaning its equivalent to cos^2x/1-sin^2x= cos^2x/sin^2x, cot^2x
the series (-1)^n-1 5n/3n^2+2 is ___ because...
conditionally converges, the function is positive and decreasing so the alternating series test applies meaning the series is convergent but then series is greater than 5(1/3x) (the function diverges by the p series test, so the original function has to diverge
the series (-1)^n-1 3/sqrt(1+n^2) is ___ convergent because
conditionally, the abs value diverges when compared to 1/n and but the series still converges because an+1 is less than an meaning an will eventually reach zero
the series (-1)^n+1 e^1/n/5n is ___ convergent because...
conditionally, the absolute value of the function (e^1/n/5n) is divergent when compared to 1/5n because it is greater than and the new function has a p-value of 1 (the function diverges), also because the function is positive and decreasing (f(n)>f(n+1)), finally since the limit of e^1/x = 1 it is seen that the limit is zero when the numerator (5n) approaching infinity
if the summation of a function is equal to 0 as its limit approaches infinity then the series will always ___
converge
the summation of 1/(k^2ln(k)+3) will ___ by the ____ test because the limit as k>infinity is = 1 and 1/k^2ln(k) converges
converge, limit comparison
1 + 1/4 + 1/9 + 1/16 + 1/25... is an example of a ___ series
convergent
the series 2+sin(n)/3^n is ___ by the ___ test
convergent, comparison (comparing it to 3/3^n using the geometric series test)
the summation of the function ke^k^2 is ___ by the ____ test
convergent, integral
the series (2/3)^n is ___ because...
convergent, the value of r will always be less than 1, and the value of n does not matter in the summation
for the p series test, when p is > than 1 the series ___
converges
the series 1 + cos(4n) ___ by the limit comparison test because it is less than or equal to 2/n^2 (p-series test)
converges
the series 5/n(ln(n)^2 __ by the integral test (u-sub)
converges
the series of 4n/n^2+3cos(2npi) is ____ because...
divergent, the value of cos(2npi) will always be 1 therefore you can use 4n/n^2+3 to test the behavior of the limit. because the limit if equal to 0, the comparison test is needed to determine that the series diverges by comparing it to the limit of 1/n that always diverges (the given function is always greater than 1/n)
by the comparison test, the series 2k^2/(8k^3-2) ___ because the series 1/k diverges
diverges
for the p series test, when p is < than or = to 1 the series ___
diverges
the series 3/n+1 ___ because it is compared to the divergence of 1/n
diverges
the series 4ln(k)/k ___ by the integral test (u-sub)
diverges
when r (of a geometric series) is greater than 1, the series ____
diverges
when the divergent series test equals 1, the series ___
diverges
the series 1/klnk+5k ___ because
diverges, by the limit comparison test the series diverges because it is less than 1/klnk that always diverges
the function 1/(xlnx) ____ and is used in what test?
diverges, comparison
the series 2n/(3n^2ln(n)+4) ___ by the ___ test
diverges, divergence
the series k+6/k^2 ___ because the function is greater than the summation of 1/k that always ____
diverges, diverges
the series n^2-36/n^2+6n ___ by the divergence test because the ___ power
diverges, highest
the series (4k/2k +1)^k ___ by the ___ test
diverges, root (the root test ends as 2>1 so it is guaranteed divergence
the series n^n/3^(2n+1) ___ by the ___ test
diverges, root test by mutiplying all exponents of the function by 1/n then taking the limit, bringing out the fraction from the bottom, for n/3^1/n the function approaches infinity
the series (-1)^n 3/(tan^-1(n))^5 ___because...
diverges, the limit of tan^-1 is equal to pi/2 and (2/pi)^5 does not equal 0 but it does say that the values oscillate without approaching 0, so the divergence test says that the series is divergent
3/5 + 4/6 + 5/7 + 6/8 + 7/9... is an example of a ___ series because the series is represented by n/n+2
diverging
m+3/(m^2lnm+2) is an example of a ____ series
diverging
what is the limit of (1+1/n)^n
e
what is the integral test?
finding the limit as x approaches infinity for the integral of 1 to x
compare the values of the series and the improper integral.
from 1 to infinity the summation will always be larger than the integral because the summation is a left reinamn sum on a dec. function. ANS: summation > integral
what is the divergence test?
if the limit of a series is any number other than 0, the series will diverge, if it does equal zero, the series may or may not converge (choose another test)
when the ratio test is = to 1, the test is ____
inconclusive
the summation of a function that starts at a value that is greater than the value that its integral starts at will be ___ than the integral because it is a ____ reinman sum
less, right
the series 3sqrt(n)/nsqrt(n)-2 diverges by the ___ test and the ___ test
limit comparison, p series
what do you do to a fraction when it is raised to the nth power and it is being multiplied to the ratio test?
multiply the fraction without the inclusion of the nth power
compute the value of lim n>infinity 4anbn/6an-3bn when lim of an=6 and bn=-2
plug in the values and divide the top and bottom, simplify
how to find the value a series converges to (geometric)?
r value/(1- r value)
to apply the ____ ____ to an infinite series of the summation of an, the value of p = lim as n>infinity (an)^1/n
root test
find all values of p for which the infinite series converges?
step 1: divide by nth power from bottom and top step 2: find the common denominator (should be a n^-#) step 3: make 1/(last step) a summation with a p series step4: multiply power by p (p must make the exponent greater than 1) step: make p equal to the negative reciprocal of the value p is being multiplied by
what is the limit of (n!)^2/(2)! (2/5)^n
steps: 1. separate the numerator 2. perform the ratio test and multiply the fraction by the entire 3. simplify like terms 4. DONT forget to multiply by the fraction
the sequence an converges and if it does, find limit when (n-4)/(n+1))^n
steps: 1. split the top and bottom into polynomials raised to the same power 2. divide n from both of the polynomials 3. use the tool lim n>infinity (1+x/n)^n = e^x to solve with the right numbers, e^-4/e^1= e^-5
rewrite the finite sum: 6/3+4 - 8/4+4 + 10/5+4 -12/6+4 +... + (-1)^6 18/9+4
summation of k=3 to 9 of (-1)^k-3 2k/k+4
the series n=1 to infinity of (-1)^n-1 n^2+3/5^n is absolutely convergent because...
the absolute value of the series converges by the ratio test
the series from n=2 to infinity of (-1)^n 4/nln(n) is conditionally convergent because
the integral test on the absolute value proves divergence and the test of convergence when taking the limit of 1/xln(x) does approach zero, meaning the series conditionally converges by the alternating series test
for the series 2k/(7ln(k)+8) the series diverges because...
the limit, as k approaches infinity, is equal to infinity
the series of the summation of n=1 to infinity of (-1)^n-1 5n/4n^2+1 is ___ because...
the numerator is inc very fast to the limit as n approaches infinity of bn, so the limit is equal to infinity, meaning the series diverge by the divergence test because (-1)^n-1 will make the values oscillate more and more when n > infinity
what is the finite sum using summation notation? 3/3+3 + 4/4+3 + 5/5+3 + 6/6+3 + ... + 9/9+3
the summation of k=3 to 9 of k/k+3
when does the alternating series test work?
when a series is positive and decreases
when -1<x<1 as n approaches infinity the limit of x^n is equal to ___ because
zero, the bottom is bigger than the top
