AP Calc BC Exam Questions and
Answers
nth term test given sigma(x) - Correct Answers -if lim(n-->inf.) x does not equal 0, then
sigma(x) DIVERGES. if lim(n-->inf.)x =0, then the test is INCONCLUSIVE
justification of nth term test - Correct Answers -"diverges by nth term test" ... NEVER
"converges by nth term test"
geometric series given sigma(x) w common ration "R" - Correct Answers -if |R| < 1 -->
converges; if |R| > 1 --> diverges
justification of geo series test - Correct Answers -"diverges bc of geometric, R=___, |R|
>1" OR "converges bc of geometric, R=___, |R|<1"
telescoping test/partial fraction decomp ONLY works if... - Correct Answers -the
denominator is factorable; if values cancel out, the leftover values are the convergence
value; if nothing cancels out, the test is INCONCLUSIVE
alternating series test given sigma(-1)^n x or sigma(-1)^n+1 x - Correct Answers -first
check that x > 0, x > x+1 (take deriv), x --> 0 (lim=0);; absolute convergence: sigma(x)
converges; conditional convergence: sigma(x) diverges, but sigma(-1)^n x converges
(on the condition that it is alternating; AST can verify convergence and divergence
watch out! alternating series test given sigma(-1)^n x or sigma(-1)^n+1 x - Correct
Answers -watch out for alternating exponent... 2n+1 is ALWAYS negative, 2n is
ALWAYS positive
does the alternating series test verify convergence or divergence - Correct Answers -
both
alternating series test justification - Correct Answers -[if conditions are met] converges
by AST; [if they are not met] diverges by AST
ratio test given sigma(xn) w nonzero terms (x sub n) - Correct Answers -if lim(n-->inf.) |
x(n+1)/xn| < 1, then the series CONVERGES; if lim(n-->inf.) |x(n+1)/xn| > 1, then the
series DIVERGES (inf. > 1); if lim(n-->inf.) |x(n+1)/xn| = 1, then the test is
INCONCLUSIVE
justification of telescoping test - Correct Answers -"converges by telescoping"
,P-Series Test given sigma(1/n^p) - Correct Answers -p>1 --> converges; p</= 1 -->
diverges; it can be ANY number over n^P
justification of P-series test - Correct Answers -"diverges by p-series test, p=__, p</= 1"
OR "converges by p-series test, p=___, p>1"
requirements of integral test given sigma(x) - Correct Answers -x must be continuous,
positive, and factorable
integral test given sigma(x) - Correct Answers -x must be continuous, positive, and
factorable; find the value of the improper integral: lim(n-->inf.) [(1 to n) x; if lim = #, then
the series converges; if lim = inf., then the series diverges
justification of the integral test - Correct Answers -"the series converges/diverges by the
integral test"
when to use the integral test - Correct Answers -use this test if you can easily
antidifferentiate; cannot use this if it doesn't meet the requirements (x must be
continuous, positive, and factorable) or if it is not antidifferentiable
limit comparison test given a positive, unknown series sigma(a); sigma(b) is a positive,
known series (p-series/geo) - Correct Answers -lim (b/a) or (a/b) = L ; if L is a positive,
real number, then either both series converge or both diverge (they behave the same);
limit cannot equal infinity or a negative #
justification of limit comparison test - Correct Answers -sigma(a) conv/div. by the limit
comparison test since sigma(b) conv/div. by _______ (p-series, geo...)
direct comparison test with sigma(a) - unknown and sigma(b) - known - Correct
Answers -CONVERGES: b>a, sigma(b) converges; DIVERGES: b<a, sigma(b)
diverges; go-to comparison series: geo + p-series; this test helps w series that would
have messy limits and integrals
benefit of direct comparison test - Correct Answers -helps w series that would have
messy limits and integrals
justification of direct comparison test - Correct Answers -since sigma(b) conv/div (geo/p-
series) and b<>a, hence it shall be known that sigma(a) conv/div by Direct Comparison
Test
series units reminders and tips - Correct Answers -pull out constants --> sigma(1/2n) =
1/2sigma(1/n) and it becomes a p-series!;;; factorials grow the fastest;;; trial and error;;;
write out justifications;;; don't forget ER+ (positive real #) for LCT;;; try simple methods
first;;; don't forget about LPT, IBP, etc.;;; if you see a factorial, think ratio test; for finding
interval of convergence, test endpoints - for finding the radius, DON'T!;;; abs. value in
ratio test and error bounds; "write the taylor series" - include sigma, "write the general
, term" - no sigma;;; 3^n / -3^n = (3/-3)^n = (-1)^n;;; |Taylor(x) - actual(x)| < error;;; can
only use taylor polynomial shortcuts when they are centered at 0
what function grows the fastest - Correct Answers -factorials
fastest growth to slowest growth - Correct Answers -n! --> 5^x --> 2^x --> x^7 --> x^2
lim(n-->inf.) 2^n/(2^n - 1) = - Correct Answers -1
if you see a factorial in a series, think: - Correct Answers -ratio test
power series (x-3)^n / 2n - Correct Answers -deriv: n(x-3)^n-n; antideriv: (x-3)^n+1 /
2n(n+1)
if you see "write the taylor series" vs "write the general term" - Correct Answers -taylor
series - include sigma; general term - don't include sigma
3^n / -3^n = (3/-3)^n = - Correct Answers -(-1)^n
taylor series -- "find f''(0)" - Correct Answers -f''(0)/2! = ______ (whatever the problem
gives you for the 2nd degree term)
you can only use taylor polynomial shortcuts when they are centered at - Correct
Answers -0
alternating series test requirements given sigma (-1)^n x - Correct Answers -x > 0, x>
x+1 (take deriv), x --> 0 (lim=0)
ratio test info - Correct Answers -don't forget absolute value; this test helps w factorials;
used to determine the intervals of convergence (x<1) --> test endpoints; simplify x terms
algebraically, then think about the limit
n! = n(n-1)!.... ; (n+7)! = (n+7)(n+6)(n+5)! - Correct Answers -
what happens if the lim <1 in the ratio test - Correct Answers -the series converges
what happens if the lim >1 in the ratio test - Correct Answers -the series diverges
what happens if the lim = 1 in the ratio test - Correct Answers -the test is inconclusive
alternating series error bound given sigma(xsubn)(-1)^n - Correct Answers -if n terms
were generated for a series, then the error would be no larger than term n+1; Rn </=
xsub(n+1); if the next term (xsubn+1) was negative in the alternating series, it is an
overapproximation; if the next term was positive, it is an underapprox; must be a
convergent alternating series
Answers
nth term test given sigma(x) - Correct Answers -if lim(n-->inf.) x does not equal 0, then
sigma(x) DIVERGES. if lim(n-->inf.)x =0, then the test is INCONCLUSIVE
justification of nth term test - Correct Answers -"diverges by nth term test" ... NEVER
"converges by nth term test"
geometric series given sigma(x) w common ration "R" - Correct Answers -if |R| < 1 -->
converges; if |R| > 1 --> diverges
justification of geo series test - Correct Answers -"diverges bc of geometric, R=___, |R|
>1" OR "converges bc of geometric, R=___, |R|<1"
telescoping test/partial fraction decomp ONLY works if... - Correct Answers -the
denominator is factorable; if values cancel out, the leftover values are the convergence
value; if nothing cancels out, the test is INCONCLUSIVE
alternating series test given sigma(-1)^n x or sigma(-1)^n+1 x - Correct Answers -first
check that x > 0, x > x+1 (take deriv), x --> 0 (lim=0);; absolute convergence: sigma(x)
converges; conditional convergence: sigma(x) diverges, but sigma(-1)^n x converges
(on the condition that it is alternating; AST can verify convergence and divergence
watch out! alternating series test given sigma(-1)^n x or sigma(-1)^n+1 x - Correct
Answers -watch out for alternating exponent... 2n+1 is ALWAYS negative, 2n is
ALWAYS positive
does the alternating series test verify convergence or divergence - Correct Answers -
both
alternating series test justification - Correct Answers -[if conditions are met] converges
by AST; [if they are not met] diverges by AST
ratio test given sigma(xn) w nonzero terms (x sub n) - Correct Answers -if lim(n-->inf.) |
x(n+1)/xn| < 1, then the series CONVERGES; if lim(n-->inf.) |x(n+1)/xn| > 1, then the
series DIVERGES (inf. > 1); if lim(n-->inf.) |x(n+1)/xn| = 1, then the test is
INCONCLUSIVE
justification of telescoping test - Correct Answers -"converges by telescoping"
,P-Series Test given sigma(1/n^p) - Correct Answers -p>1 --> converges; p</= 1 -->
diverges; it can be ANY number over n^P
justification of P-series test - Correct Answers -"diverges by p-series test, p=__, p</= 1"
OR "converges by p-series test, p=___, p>1"
requirements of integral test given sigma(x) - Correct Answers -x must be continuous,
positive, and factorable
integral test given sigma(x) - Correct Answers -x must be continuous, positive, and
factorable; find the value of the improper integral: lim(n-->inf.) [(1 to n) x; if lim = #, then
the series converges; if lim = inf., then the series diverges
justification of the integral test - Correct Answers -"the series converges/diverges by the
integral test"
when to use the integral test - Correct Answers -use this test if you can easily
antidifferentiate; cannot use this if it doesn't meet the requirements (x must be
continuous, positive, and factorable) or if it is not antidifferentiable
limit comparison test given a positive, unknown series sigma(a); sigma(b) is a positive,
known series (p-series/geo) - Correct Answers -lim (b/a) or (a/b) = L ; if L is a positive,
real number, then either both series converge or both diverge (they behave the same);
limit cannot equal infinity or a negative #
justification of limit comparison test - Correct Answers -sigma(a) conv/div. by the limit
comparison test since sigma(b) conv/div. by _______ (p-series, geo...)
direct comparison test with sigma(a) - unknown and sigma(b) - known - Correct
Answers -CONVERGES: b>a, sigma(b) converges; DIVERGES: b<a, sigma(b)
diverges; go-to comparison series: geo + p-series; this test helps w series that would
have messy limits and integrals
benefit of direct comparison test - Correct Answers -helps w series that would have
messy limits and integrals
justification of direct comparison test - Correct Answers -since sigma(b) conv/div (geo/p-
series) and b<>a, hence it shall be known that sigma(a) conv/div by Direct Comparison
Test
series units reminders and tips - Correct Answers -pull out constants --> sigma(1/2n) =
1/2sigma(1/n) and it becomes a p-series!;;; factorials grow the fastest;;; trial and error;;;
write out justifications;;; don't forget ER+ (positive real #) for LCT;;; try simple methods
first;;; don't forget about LPT, IBP, etc.;;; if you see a factorial, think ratio test; for finding
interval of convergence, test endpoints - for finding the radius, DON'T!;;; abs. value in
ratio test and error bounds; "write the taylor series" - include sigma, "write the general
, term" - no sigma;;; 3^n / -3^n = (3/-3)^n = (-1)^n;;; |Taylor(x) - actual(x)| < error;;; can
only use taylor polynomial shortcuts when they are centered at 0
what function grows the fastest - Correct Answers -factorials
fastest growth to slowest growth - Correct Answers -n! --> 5^x --> 2^x --> x^7 --> x^2
lim(n-->inf.) 2^n/(2^n - 1) = - Correct Answers -1
if you see a factorial in a series, think: - Correct Answers -ratio test
power series (x-3)^n / 2n - Correct Answers -deriv: n(x-3)^n-n; antideriv: (x-3)^n+1 /
2n(n+1)
if you see "write the taylor series" vs "write the general term" - Correct Answers -taylor
series - include sigma; general term - don't include sigma
3^n / -3^n = (3/-3)^n = - Correct Answers -(-1)^n
taylor series -- "find f''(0)" - Correct Answers -f''(0)/2! = ______ (whatever the problem
gives you for the 2nd degree term)
you can only use taylor polynomial shortcuts when they are centered at - Correct
Answers -0
alternating series test requirements given sigma (-1)^n x - Correct Answers -x > 0, x>
x+1 (take deriv), x --> 0 (lim=0)
ratio test info - Correct Answers -don't forget absolute value; this test helps w factorials;
used to determine the intervals of convergence (x<1) --> test endpoints; simplify x terms
algebraically, then think about the limit
n! = n(n-1)!.... ; (n+7)! = (n+7)(n+6)(n+5)! - Correct Answers -
what happens if the lim <1 in the ratio test - Correct Answers -the series converges
what happens if the lim >1 in the ratio test - Correct Answers -the series diverges
what happens if the lim = 1 in the ratio test - Correct Answers -the test is inconclusive
alternating series error bound given sigma(xsubn)(-1)^n - Correct Answers -if n terms
were generated for a series, then the error would be no larger than term n+1; Rn </=
xsub(n+1); if the next term (xsubn+1) was negative in the alternating series, it is an
overapproximation; if the next term was positive, it is an underapprox; must be a
convergent alternating series
