2

Using L’Hospitals Rule find the following limits lim n2 e−n = 0, lim (lnnn) = n→∞ n→∞ √ √ 0, lim n sin(π/ n) = π. n→∞

If f (x) is a continuous function, and lim an = L, then lim f (an ) = f (L). n→∞

n→∞

2nπ Using this find the following limits lim e1/n = e0 , lim tan( 1+8n ) = tan(π/4), n→∞ n→∞ q √ 3 −1 n 1 21+3n = lim 3 27n3n+9n 2 +1 = 3 , lim cos(2/n) = cos(0), lim arctan 2n = π/2, lim

n→∞ 3

n→∞

n→∞

n→∞

2 .

√ (ln n)2 + n −2n4 +n2 +1 en +e−n 3 +1+n , lim 3en +ln n , lim 2 +1 . 3n (ln n) n→∞ n→∞ n→∞ biggest thing in numerator −2n4 +n2 +1 We take the lim biggest . For example lim = 3 thing in the denominator n→∞ n→∞ 3n +1+n √ 4 n −n n (ln n)2 + n −2n e +e e lim −2n = lim = limit does not exist. lim = lim =1/3, lim = 3 n n 2 n→∞ 3n n→∞ 3 n→∞ 3e +ln n n→∞ 3e n→∞ (ln n) +1 √ n lim 2 = 0(By LHospital’s Rule). n→∞ (ln n)

To Determine the limit of fractions like lim

Remember/Recall the hierarchy of bigness different kind of functions . loga√ among √ rithmic functions(ln n, (ln n)2 , etc) N . that is eventually increasing or eventually decreasing. Then lim an converges and satisfies n→∞ m ≤ lim an ≤ M . n→∞

using MST find the following limits, — suppose a1 = 1, an+1 = 3− a1n , and assume (donot prove) this sequence is increasing and an < 3 for all n, deduce lim an converges and find the limit. Ans

n→∞

√ 3+ 5 2

1 , and assume (donot prove) this sequence is desuppose a1 = 2, an+1 = 3−a n creasing and 0 ≤ an ≤ 2 for all n, deduce lim an converges and find the limit. Ans n→∞

√ 3− 5 2

Suppose {an }, {bn } are two sequences, and lim an converges, and lim bn n→∞

n→∞

converges. Then lim (an bn ), and lim an + bn converge. and lim (an bn ) = n→∞

n→∞

n→∞

( lim an )( lim bn ), and lim (an + bn ) = lim an + lim bn . n→∞

n→∞

n→∞

n→∞ −n

lim (cos(1/n) arctan(n)) = π/2, lim (e1/n + e

n→∞

n→∞

n→∞

For example

)=1+0=1

Suppose {an }, {bn } are two sequences, such that lim an converges and lim an 6= n→∞

n→∞

0, but lim bn diverges. Then lim (an bn ), and lim an + bn diverges. For exn→∞

n→∞

n→∞

n diverges because lim n+1 = 1 6= 0 converges, but lim cos(n) ample lim n cos(n) n→∞ n→∞ n→∞ n+1 diverges.

Using this determine if the following limits converge lim arctan(n) sin(1/n) = 0, n→∞

2 lim cos(n)n 2 n→∞ 3n +1

diverges, limits converge.

2 n lim cos 2 n→∞ n +1

n −n

= 0, lim e e n→∞

= 1. WATCH OUT the last two

The following is a flowchart/steps one can use to determine the convergence of a series

3

start

sn is given or can find sn i.e telescoping series

find lim sn n→∞

lim sn =PL converges, an converges and =L

limP sn diverges, then an diverges find a by plugging in the starting value of n into the formula, and |r|

P

an is a geometric series

|r| < 1, then P an converges a and = 1−r

|r| > 1 then P an diverges if lim |an | 6= 0, TOD implies P an diverges

TOD:is lim |an | = 0 ?

lim | an+1 | n→∞ an then

P

> 1,

an diverges

| < 1, lim | aan+1 Pn then an converges absolutely

lim |an | = 0, TOD is inconclusive, apply ratio test, find lim | aan+1 | n

n→∞

n→∞

| = 1, lim | aan+1 n then Ratio Test is inconclusive, Is this an alternating series n→∞

If this is an alternating series, is lim bn = 0

lim bn = 0, and {bn } is eventually decreasing, then PAST implies an converges

if lim bn = 0, then is {bn } eventually decreasing no

is

P

an a positive series, i.e is an ≥ 0 for all n

P is |an | convergent(use Comparision test or limit Comparision test)

lim 0, then Pbn 6= n−1 (−1) bn diverges if an approxto Pimation (−a)n−1 bn is reqd, then find n such that bn+1 ≤ maxerror, then by ASET sn is a good enough approximation P |an |P converges, then an converges absolutely, hence converges

P

|anP | diverges, then an may still converge

Apply Limit Comparision Test

LCT doesnot help, then use Comparison Test

try integral Test

4 ∞ P

Test of Divergence(TOD): Let

an be a series. If lim |an | 6= 0, then n→∞

n=1

∞ P

an diverges. if lim |an | = 0, then the test is inconclusive. This is slightly n→∞

n=1

∞ P

different from the book, but is easier to apply in the cases like

n=1

Using TOD, determine if the following series converge. diverges,

∞ P

n−1 3n−1

n=1 ∞ P

verges,

diverges,

∞ P

n(n+1) (n+2)2

n=1 ∞ P

arctan(n) diverges,

n=1

n=1

∞ P

diverges,

(−1)n n n+1

∞ P

(−1)n (n2 −1) n2 +1

cos(n) diverges,

n=1

cos(1/n) diverges,

n=1 ∞ P

diverges,

n=1

∞ P

n=1 ∞ √ P n

2n 3n+1

2 di-

n=1 (−1)n n 3n2 +n+1

inconclusive

a A geometric series a + ar + ar2 + · · · converges if |r| < 1, then the sum is 1−r , it diverges if |r| ≥ 1. a is called the first term, r is called the common ratio. For the ∞ n+1 P 7 series 10n . the starting value of n is 3, plugging this into the formula we get n=3

(74 /103 ) 74 103 , and r = 7/10. since |r| < 1 we get sum is 1−(7/10) . while ∞ P 3n series 2n+1 , we have |r| = 3/2 > 1, hence this series diverges. n=1

the first term a = the geometric

in

Determine if the following geometric series converge. If they do, then find their sum. ∞ ∞ ∞ P P P 12 10n √1 , 4 + 3 + 9/4 + 27/16 + · · · , 10 − 2 + 0.4 − 0.08 + · · · , , , n n+1 (−5) (−9) ( 2)n

n=1

n=1

n=1

1 + 0.4 + 0.16 + 0.064 + · · · ,

∞ P n=1

∞ P

Let

an ,

n=1

∞ P

cos(1)k .

n=1

bn be two series then if one of the two series converges, while

n=1

the other one diverges then ∞ P

verge, then

∞ P

6(0.9)n−1 ,

n=1

what happens to

∞ P

(an + bn ) diverges. If both the series con-

n=1

(an + bn ) converges. If both the series diverges, we donot know ∞ P

(an + bn ). In class, we denoted this by C + C = C, C + D =

n=1

D, D + D =? using this determine if the following series converge diverges, ∞ P

∞ P

n=1 n−1

[(0.8)

n=1

1 n(n+1) )

1+2n 3n

= 1/2 + 2,

∞ P n=1

1+3n 2n

n

− (0.3) ](find the sum, leave your answer as a fraction),

n=1

( e1n +

∞ P

converges,

∞ P n=1

( n2 +

3 5n )

Find all values of c for which the series

diverges ∞ P

( nc −

n=1

only for c = 1, diverges for all other values of c.

1 n+1 ),

converges.Ans- Converges

5 ∞ P

(v.important) Given a series

an , sn is the sum of the terms from the start

n=1

upto and including the term one gets by pluging in n into the formula of the series. ∞ P 1 For example for the series ln n , s3 = 1/ ln 2 + 1/ ln 3. n=2

if the series starts at n = 1, i.e for example if the series is

∞ P

∞ P

an , then a1 = s1 but for n > 1, an = sn − sn−1 .

n=1

an , and it is given sn =

n=1

n−1 n+1 ,

then a1 = s1 = 0, while

a3 = s3 − s2 = 2/4 − 1/3 = 1/6. ∞ P n=1 ∞ P

an converges if and only if lim sn converges. and if they do converge then n→∞

n−1 n→∞ n+1

an = lim sn . For example in the above example lim sn = lim n→∞

n=1

hence

∞ P

n→∞

= 1.

an converges and equals 1.

n=1

Suppose the nth partial sum of the series n−2 2n .

a1 = 1/2, an = ∞ P

determine if

∞ P

an is sn = 3 − n2−n . Then find

n=1

∞ P

an converges yes. If it does find the value of

n=1

an = 3.

n=1

(v.important) Using the telescoping series method find sn for the following series. then using sn determine if the series converges. if it does then ∞ ∞ P P 1 2 find the value of the sum. ( √1n − √n+1 ) = 1 converges, n2 +4n+3 = 7/12 converges,

∞ P n=1

n=1

1 n(n+1)

= 1 converges,

∞ P n=1

n=2

n ln( n+1 )

diverges.

Integral Test, Comparision Test, Limit Comparision Test, can be used only for series with positive terms. if an ≤ bn , and

∞ P

bn converges, it is NOT always true that

n=1

∞ P

an converges. for

n=1

example consider the series (−1 − 1 − 1 − 1 − 1 − · · · ) < (0 + 0 + 0 + 0 + · · · ). we cannot use comparision test because an is not positive. the p-series ∞ P

∞ P n=1

1 np ,

converges if p > 1 diverges if p ≤ 1.

For example

n.1 diverges as p = −.1 ≤ 1.

n=1

using the p-series test determine if the following series converge. ∞ P n=1

2 n0.85

diverges,

∞ P

(n−1.4 + 3n−1.2 ) converges,

n=1

∞ P n=1

∞ √ P 5

n diverges,

n=1

typo, 1 +

1 √ 2 2

+

1 √ 3 3

+ ···

6

converges Using the integral test determine if the following series converge

∞ P

ne−n converges,

n=1

∞ P n=2

1 n ln n .

diverges

Given a series like

∞ P

an =

n=1 ∞ P

need a second series

∞ P n=1

en +n+1 n2 en +1 .

to apply limit comparison test, to this we

an = 1 or n→∞ bn n biggest summand in the numerator of an e 1 biggest summand in the denominator of an = n2 en = n2 .

bn , which is easier to handle but such that lim

n=1

any nonzero number. Take bn = Then it is clear

en +n+1 n2 en +1 1 n→∞ n2

lim

. we know

∞ P n=1

1 n2

=1

converges, hence the given series must converge by limit compar-

ison test. ∞ P

using the limit comparison test determine if the following series converge D,

∞ P n=1

n 2n3 +1

C,

∞ P n=1

n=1 n3 n4 −.5

D, 1 + 1/4 + 1/7 + 1/10 + · · · D (find a formula for

the nth term, then use LCT,

∞ P n=1

1 n2 +9

∞ P

an , if the series

n=1

converges absolutely. In this case

∞ P

C,

1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + · · · D, Given a series

√1 n+4

n2 −1 3n4 +1

n=1 ∞ P 1+4n 1+3n n=1

∞ P n=1 ∞ P

C,

D,

∞ P

n−1 n4n

n=1 ∞ P √ 1 n3 +1 n=1

C,

∞ P

C,

n=1 ∞ P n=1

4+3n 2n

D,

sin( n1 ) D

|an | converges, then we say

∞ P

an

n=1

an must also converge. Hence ab-

n=1

solute convergence implies convergence. But the converse is false. it may happen that a series converges, but it does not converge absolutely. If the terms of a series are not positive. Then we cannot apply comparison, LCT or integral test to it. In this cases sometimes it helps to check the series for absolute ∞ P convergence. to do that we check that the series |an | is convergent. if it is so, then

∞ P

n=1

an is absolutely convergent, hence convergent. But if Ser1an is not

n=1

absolutely convergent, it may still be simply convergent. For example

∞ P n=1

convergent, but not absolutely convergent. using absolute convergence show that the series

∞ P n=1

sin(n) n2

is convergent.

(−1)n n

is

7

in using comparison test between two positive series, if the Bigger series converges, then the smaller series converges. If the smaller series diverges, then the bigger series diverges. (important)To determine convergence of the series parison test directly to compare the series with

∞ P n=1

∞ P n=1

1 4n ,

sin(4n) 4n ,

we cannot use com-

sin(4n) 4n

because

is not positive

for all n. instead we use a combination of absolute convergence and comparison ∞ ∞ P P 1 1 | sin(4n) test as follows | sin(4n) 4n | < 4n , since 4n converges, we must have 4n | n=1

n=1

converges by comparison test. Hence the given series converges absolutely, hence must also converge. Use comparison test, or a combination of absolute convergence and comparison ∞ P sin(4n ) absolutely test, if required, to determine if the following series converge. 4n Converges, lutely C.,

∞ P

cos2 n n2 +1

n=1 ∞ P sin2 n n3 n=1

Absolutely C.,

n=1

∞ P n=1

2+(−1)n √ n n

Absolutely C.,

∞ P

n=1

1+sin(n) 10n

Abso-

Absolutely C.

Alternating Series Test Let bn > 0 and

∞ P

(−1)n−1 bn be an alternating se-

n=1

ries. Then if the sequence of absolute values {bn } is eventually decreasing, that is decreasing for n > N , for some number N, and lim bn = 0. Then the altern→∞ nating series converges. If lim bn 6= 0, Then by TOD the alternating series n→∞

diverges. If the sequence {bn } is not eventually decreasing then the alternating series Test is inconclusive. Using the AST determine if the following series converges. ∞ P n=1

D,

1 (−1)n−1 ln(n+4) C, ∞ P

(−1)n−1 enn C,

n=1

∞ P

(−1)n−1 3n−1 2n+1 D,

n=1 ∞ P

n=1

n=1

∞ P

2

(−1)n−1 8nn C,

∞ P

(−1)n−1 √nn3 +2 C,

∞ P

(−1)n−1 √1n C,

n=1 ∞ P

(−1)n−1 cos(π/n)

n=1

n (−1)n−1 arctan C. n2

n=1

To check that the sequence {bn } is eventually decreasing, we have two methods. in ∞ P first case it might be obvious that, it is so. for example in the case (−1)n−1 n1 . n=1

in these cases it is enough to just say so. If it is not obvious for example in ∞ P 2 (−1)n−1 n3n+1 . we use the first derivative test in the interval (1, ∞) as fol-

n=1

x2 x3 +1 .

Then f 0 (x) =

x(2−x3 ) (x3 +1)2 .

in the interval (1, ∞) the factor √ x/(x + 1) > 0, but (2 − x ) < 0 for x > 2. hence for x > 3 2, f 0 (x) < 0. hence the sequence {bn = f (n)} is decreasing for n ≥ 2. lows. Let f (x) = 3

2

3

√ 3

8 2

Prove that the sequences { enn }, { 2nn }, are eventually decreasing. (Very Important) Alternating Series Estimation Theorem Let

∞ P

(−1)n−1 bn

n=1

be an alternating series which satisfies the conditions of the alternating series Test, ∞ P and hence converges. Let s = (−1)n−1 bn . Let sn be the partial sum upto n=1

and including the nth term (−1)n−1 bn , Then the error in approximating s by ∞ P sn =|s−sn | ≤ bn+1 . For example let s = (−1)n−1 n1 . Then |s−(1−1/2+1/3)| ≤ n=1

1/4. Using the ASET, for each of the following series find n such that, |s − sn | ≤ .001. ∞ ∞ ∞ ∞ ∞ P P P P P (−1)n−1 ln1n , (−1)n−1 8nn , (−1)n−1 n16 , (−1)n−1 3n1n! . (−1)n−1 n1 , n=2

n=1

n=1

n=1

n=1

Ratio Test is generally easier to apply than the alternating series Test. for example ∞ ∞ P P 2 in the cases (−1)n−1 enn , (−1)n−1 8nn , checking that the {bn } is eventually n=1

n=1

decreasing is cumbersome, while the ratio test is straight forward. while applying ratio test, we have to simplify ratios like 1·2···(2n) (2n)! 1 = 1·2····(2n)·(2n+1)·(2n+2) = (2n+1)(2n+2) , n + 1, (2(n+1))! 1·2···(2n+1) 1·2····(2n+1)·(2n+2)·(2n+3) ∞ P

Given a series

=

(n+1)! = 1·2···n·(n+1) n! 1·2···n (2n+1)! (2n+1)! = (2(n+1)+1)! (2n+3)!

= =

1 (2n+2)(2n+3)

an , an = the term we get when we substitute n into the formula

n=1

of the series. For example in the series

∞ P n=0

(−1)n x2n+1 (2n

+ 1)!, a3 =

(−1)3 x2.3+1 (2.3

+ 1)!.

see that a3 is the 4th term from the start. ∞ P | and Given a series an , to apply ratio test to this series, we first find | aan+1 n n=1

simplify it. . Then we take it’s lim | aan+1 |. If lim | aan+1 | < 1, the series conn n n→∞

n→∞

verges absolutely. If lim | aan+1 | > 1, the series diverges. If lim | aan+1 | = 1, n n n→∞ n→∞ then the Ratio test is inconclusive. When it is inconclusive, we should try AST, LCT, CT, if the series has negative terms then try a combination of absolute convergence and comparison test, IT, in that order. Apply ratio test to the following series.

∞ P

3

(−1)n 3nn Absolutely C.,

n=1

hard question, don’t worry if you can’t do this), Absolutely C., lutely C.,

∞ P

∞ P n=1

1 3n n!

n (−1)n 2n4 n=1

Absolutely C.,

D,

∞ P n=1

∞ P n=1

10n (n+1)42n+1

∞ P

n=1

(−1)n n13 inconclusive,

n=1 (−3)n n3

∞ P

D,

∞ P

n! 100n

n=1 ∞ P

Absolutely C.,

D,

∞ P n=1

n! (−2)n (2n)! n=1

nn n! ∞ P

D(is a 2

n (−1)n 10 n

n=1 k

k 23k AbsoAbsolutely

9

C.,

∞ P n=1

√

n 1+n2

Inconclusive

A power series is a function which is represented as a series. the series has a variable x in it. for some values of x, the power series converges, while for the remaining values of x,the power series diverges. The set of values of x for which the power series converges is always an interval of one of the following types ( ),( ], [ ], [ ). It is called the interval of convergence. often denoted by IOC. the interval of convergence is also the domain of the function represented by the given power series. the radius of this interval is called the radius of convergence. often denoted by R. the center of this interval is called the center of convergence. often denoted by a. Every power series converges in it’s center. So the interval of convergence is never empty. the interval of convergence is always symmetric about the center. Suppose the interval of convergence is (1, 2], then the center is (2 + 1)/2 = 3/2, while the radius is (2 − 1)/2 = 1/2. the ”power” in the word ”power series” comes from the fact that the terms of the series contain a power of x − a, where a is the center of the power series. A power series converges absolutely in every open interval inside the interval of ∞ P (x−1)n has interval of convergence convergence. For example the power series n n=0

[0, 2). It converges absolutely in (0, 2). but at the boundary point 0, it doesnot converge absolutely. Outside the interval of convergence, a power series diverges. by the way the center of C. of this series is 1, and the radius of C. of this series is 1. Let

∞ P

cn (x − a)n , be a power series. then there are three types of radius of con-

n=0

vergence. if the IOC is [a, a]. the radius is 0. in this case the power series converges ∞ P only at it’s center. for example n!(x − 1)n has IOC=[1, 1]. if the interval is n=0

of the type (a − R, a + R], [a − R, a + R], [a − R, a + R), (a − R, a + R), for some ∞ P nonzero finite R, for example for the series (x − 1)n , which has IOC=(0, 2), n=0

hence has radius=1. The third type is when the IOC=(−∞, ∞), in which case we ∞ P (x−1)n . say the radius of convergence is ∞ for example in the case n! n=0

Find the radius of convergence, and the interval of convergence of the following ∞ ∞ ∞ √ P P P n (−1)n xn √x power series. ,R=1, IOC=[-1,1), R=1, IOC=(-1,1], n(x)n n+1 n+1 n=0

R=1, IOC=(-1,1), ∞ P n=1

∞ P n=0

n=0

(x)n n!

(−1)n n2 xn R=2,IOC= 2n

IOC=[0,0],

∞ P n=0

R=∞ IOC=(−∞, ∞),

(-2,2),

(−1)n (x−3)n 2n+1

∞ P n=0

∞ P n=1

(−1)n x2n (2n)!

R=1 IOC=(2,4],

n=0

(10x)n n3

R=1/10,IOC=[-1/10,1/10],

R=∞ IOC=(−∞, ∞), ∞ P

n=0

∞ P n=0

(3x−2)n n3n

(2n)!xn 2n

R=1 IOC=[-1/3,5/3),

R=0 ∞ P

n=0

n!(2x−1)n 2

10

R=0 IOC=[1/2,1/2],

∞ P n=1

xn 1·3·5···(2n−1)

R=∞ IOC=(−∞, ∞).

Very Important for exam {c0 , c1 , . . .} is a sequence such that but

∞ P

∞ P

cn 5n converges

n=0

cn (−1)n 8n diverges. what can you say about the convergence of the fol-

n=0

lowing series ?

∞ P

cn Absolutely C.,

n=0 ∞ P

maynot converge,

∞ P

cn (−1)n Absolutely C.,

n=0

cn (−1)n 10n Diverges,

n=0

may or maynot converge,

∞ P

∞ P

∞ P

cn 6n n=0 ∞ P

cn 3n Absolutely C.,

n=0

may or

cn (−1)n 7n

n=0

cn xn . Absolutely C. for x in (−5, 5), Diverges for

n=0

|x| > 8. converges at x = 5, diverges at x = −8.