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Chapter 1 NUMERICAL SERIES
1. Sequences……………………………………………………. 1 2. Definitions. General Criteria………………………………… 17 3. Series with Nonnegative Terms……………………………. 28 4. Approximate Computation of Sums……………………….. 38 5. Improper Integrals and Series……………………………… 42 6. Infinite Products……………………………………………… 59
7. SOLVED PROBLEMS
Exercise 38. (a) 1lnlim =∞→
n
nn (b) ∞=
∞→
n
nn!lim
(c) 1!lnlim =∞→
n
nn
(d) e
nnnnn
nn
4))...(2)(1(1lim =+++∞→
(e) nn
kn k0sinlim
1
=∏=
∞→
π
(f) 0)!2()2(lim =
∞→ nn n
n (g) 0
)!2()!(lim
2
=∞→ n
nn
(h) 1ln
!lnlim =⋅∞→ nn
nn
(i) ∞=+++∞→
n nnn
nn...21lim (j) ∑
=∞→
=n
kn kn 1211lim
(k) ∞=+∏=
+
∞→
n
k
kn
nkn
1
lim
(l) ∑=
∞→
−=
+
n
kn knk
nn 1 3)22(21lim
Solutions: (a) Using Cauchy’s criterion we have:
1ln
)11ln(lim1
ln
)11ln(lnlim
ln)1ln(limlnlim =
++=
++=
+=
∞→∞→∞→∞→ nn
nn
n
nnn
nnn
n
n
.
64
(b) Analogously:
∞=+=+
=∞→∞→∞→
)1(lim!
)!1(lim!lim nn
nnnn
n
n.
(c) Applying root’s criterion, then Stolz-Cesaro’s criterion we obtain:
.1)1ln(
12ln
lim1!ln)!1ln(
)1ln()2ln(lim1
!ln)1ln(lim1
!ln)1ln(!lnlim
!ln)!1ln(lim!lnlim
=+++
+=−+
+−++
=+
+=++
=+
=
∞→∞→
∞→∞→∞→∞→
nnn
nnnn
nn
nnn
nnn
nn
nnn
n
n
(d) Again, Cauchy’s criterion implies:
( ) .4
)11(
1lim41
lim4)1()1(
)22)(12(lim
)(...)2)(1()1()22)(12)((...)3)(2(limlim
1
11
en
nn
nnnnn
nnnnnnnnnnnn
nknl
nn
n
nn
n
n
n
n
nn
n
kn
=+
=+
=++⋅++
=
=+++⋅+
⋅+++++=
+=
∞→∞→+∞→
+∞→=
∞→ ∏
Or, because:
∑=
∏= ==
+
∞→
+
∞→
n
k
nn
k nk
n
n
nk
neel 11
)1ln(1)1(ln
limlim
denoting
[ ]nkxIRfxxf kk
k ==→+= ξ,1,0:),1ln()( ,
we have: ∑∑=
−=
−=+n
kkkk
n
k
xxfnk
n 11
1
))(()1ln(1 ξ ,
This is a Riemann sum for the continuous, hence integrable function f, therefore:
65
.4,4ln12ln2
)1ln()1()1ln()'1(
)1ln()()1ln(1lim
4ln1
0
1
0
1
0
1
0
1
01
eeland
ex
xxdxxx
dxxdxxfnk
n
e
n
kn
===−=−
−++=++=
=+==+
∫
∫∫∑=
∞→
(e) 01
sinlimsin
sinlimsinlim
1
1
1
1
=+
==∞→
=
+
=
∞→=
∞→
∏
∏∏ n
k
kk nn
k
n
k
nn
n
kn
ππ
ππ .
(f) Let: )!2()2(
nna
n
n = ; then:
(1) 1312
)11(12
1)2(
1)12)(22()22)(22(
)2()!2(
)!22()22( 1
1
<≤+
<++
=
=⋅++++
=⋅++
=+
+
ene
nn
nnnnn
nn
nn
aa
n
n
n
n
n
n
n
for all ; it follows that is a decreasing sequence of positive numbers, hence it converges, i.e. there exists such that . But, from (1):
*IN∈n )( na0≥a
nnaa
∞→= lim
00)3(33 1 ≤⇔≤−⇔≤⇒≤∞→+ aaeeaaeaa
nnn , hence . 0=a
(g) Let )!2(
)!( 2
nnan = . Then:
)12(2
1)12)(22(
)1()!()!2(
)!22(])!1[( 2
2
21
++
=++
+=⋅
++
=+
nn
nnn
nn
nn
aa
n
n ;
but: *1243331
)12(21 IN∈⇔≤⇔+≤+⇔≤++ nnnnn
n ,
hence: )(..,,131 *1
nn
n aeina
a IN∈∀<≤+ is a decreasing sequence of
positive numbers; therefore it is convergent. Let: 0lim ≥=∞→ nn
aa . But
, and, for , we obtain 03 1 ≤−+ nn aa ∞→n 03 ≤− aa or . Consequently
0≤a0=a .
66
(h) We have , hence we can apply Cesaro-Stolz’ criterion and we obtain:
∞→nn ln
.11)11ln(
)1ln(
1lim)1ln()11ln(
)1ln(lim
)1ln(]ln)1[ln()1ln(lim
ln)1ln()1(!ln)!1ln(lim
ln!lnlim
=++⋅
+
=+++
+=
=++−+
+=
−++−+
=
∞→∞→
∞→∞→∞→
nnn
nnn
nnnn
nn
nnnnn
nnnnn
nnnn
n
(i) Applying Cauchy’s criterion then Cesaro-Stolz’ criterion we obtain:
.)2(lim)2()1
11(lim)2()12(lim
)1()2(lim
...21)1(...21lim...21lim
11
1
21
∞=+=++
+=+++
=
=++
=+++++++
=+++
∞→
+
∞→
+
∞→
+
+
∞→
+
∞→∞→
nenn
nnn
nn
nnn
n
n
n
n
n
n
n
nnnn
nnn
nn nnn
n
(j) Cesaro-Stolz’ criterion implies (since ∞→n ):
.2)1
1(lim)1(1
1lim1
11
lim
1
lim 1 =+
+=−+⋅+
++=
−++=
∞→∞→∞→
=
∞→
∑n
nnnn
nnnn
nn
knnn
n
k
n
(k) We observe that:
,))...(2)(1[(
2...212...21
2
222221
1
n
nnnnnnn
k
knn
nnnn
nnnnnnkna
+++>
>⋅⋅+⋅+>⋅⋅+⋅+=+= ++
=
+∏
hence, from Cauchy’s criterion:
;)]12(2[lim]))...(2)(1(
)22)(12)()...(2([limlim 21
21
∞=+=+++
++++≥
∞→∞→∞→n
nnnnnnnnna
nnnn
consequently: ∞=∞→ nn
alim .
(l) We remark that:
67
∑ ∑ ∑= = =
−−=+
=+
=n
k
n
k
n
kkkkn xxf
nk
nk
nknk
nns
1 1 11 ))((
1
11 ξ ,
where x
xxfRfnknkx kk
+=→===
1)(,]1,0[:,,...,1,0,ξ ; hence
is a Riemann sum for the function ; but is a continuous function and:
ns
f f
.)224(3123
222322)1(2
3)1(2
)1
11(111
1)(lim
212
31
02
123
1
0
1
0
1
0
1
0
−=+−⋅−⋅
=+−+
=
=+
−+=+−+
=+
== ∫∫∫∫∞→
xx
dxx
xdxx
xdxxxdxxfsnn
Exercise 39. Test the series for convergence using the
definition:
(a) ZIRaanann
\,)1)((
10
∈+++∑
∞
=
(b) ∑∞
=
+
0 21
nn
n
(c) ∑∞
= ++02 1
1arctann nn
(d) ∑∞
=++ +⋅−1
1212 4209520
nnnn
n
(e) ∑∞
= −++−−
122
234
)14)(1(116816
n nnnnnn
(f) ∑∞
=⎥⎦
⎤⎢⎣
⎡+−
−+
12 1)1(2
)12(21lnn n
n .
Solutions. (a) The general term can be decomposed as:
111
)1)((1
++−
+=
+++=
ananananan ,
and the partial sum of n-th order:
aanaakakakaks
n
k
n
kn
11
111
11)1)((
100
→++
−=++
−+
=+++
= ∑∑==
,
therefore the series is convergent and it’s sum:
aanann
1)1)((
10
=+++∑
∞
=
.
68
(b) For the partial sum of n-th order: ∑=
+=
n
kkn
ks0 2
1 , we
consider the derivative of sum of a geometric progression: [ ]
2
212
0 0
1
)1()1()2(1)'
1()'()1(
xxxxxn
xxxxxk
nnnn
k
n
k
kk
−−+−+−
=−−
==++++
= =
+∑ ∑ ,
and, for 21
=x we obtain:
4)21
21(4
21
21
21)
221(4 21 =+→⎥⎦
⎤⎢⎣⎡ −+
+−= ++ nnn
ns ,
therefore is convergent, i.e. our series converges and its sum is
4; we write:
ns
42
10
=+∑
∞
=nn
n .
(c) Since:
IN∈−+=++−+
=++
kkkkkkk
kk,arctan)1arctan(
)1(1)1(arctan
11arctan 2 ,
the partial sum
421arctan)1arctan(
11arctan
12
ππ−→−+=
++= ∑
=
nkk
sn
kn ,
and we conclude that our series is convergent and its sum is 4π .
(d) If we mark and then the general term is: na 5= nb 4=
baa
baa
babaab
baba
abbaba
aban 455
)45)(()54)((5495 22 −
−−
=−−
=−−
=+−
=
hence the partial sum of n-th order is:
415)
54(1
15
4555
455
455
1
111
1
111
=−→−
−=
=−
−=−
−−
==
+
=++
+
=++∑ ∑
n
n
knn
nn
kkk
k
kk
k
kn as
;
therefore the series is convergent and: 4)45)(45(
201
11 =−−∑
∞
=++
nnnnn
n
.
69
(e) The general term can be decomposed in simple
fraction: na
222
234
)12(12121)12()12)(1(116816
++
++
−+
++=
+−++−−
=n
EnD
nC
nB
nA
nnnnnnnan ,
and, after calculations, we obtain:
;0,1
11411
,13
1411,19
116816,1
===−
+−+=
−=+−−
=−=−
+−+==
DCF
EBA
therefore:
0)12(
111
11)12(
1)12(
11
11222
11
→+
+−+
−=+
+−
−+
−== ∑∑== nnkkkk
asn
k
n
kkn
consequently the series is convergent and its sum is 0.
(f) The partial sum of n-th order:
∞→+=+−
+=
+−−++−
= ∏∑==
)12ln(1)1(2
12ln1)1(2
241)1(2ln 2
12
2
12
2
nk
kk
kksn
k
n
kn
when ∞→n and the series is divergent.
Exercise 40. Check whether the necessary condition for convergence is fulfilled:
(a) ...161
111
51
+++ (b) ...167
115
53
+++
Solutions. (a) The general term of our series:
015
1→
+=
nan when , hence the necessary condition for
convergence is verified; however:
∞→n
),0(51
151∞∈→
+=
nn
n
an , hence,
the series has the same nature with harmonic series and, therefore is divergent.
(b) Here 052
1512
≠→++
=nnan , hence the series diverges.
70
Exercise 41. Test the series for convergence by means of
the integral test:
(a) ...43
32
21
333 +++ (b) ...31
321
211
1222 +
++
++
+
(c) ...4ln4
13ln3
12ln2
1222 +++
Solutions. (a) The general term is:
)()1( 3 nf
nnan =+
= , where [ ] +→∞+
= IR,1:,)1(
)( 3 fx
xxf ,
and the generalized integral:
[ ]
;41
41
21
)1(21
11
)1()1()1(11
)1()(
12
1 1
323
1 31
∞<=−=+
++
−=
=+−+=+−+
=
=+
=
∞
∞ ∞ −−
∞∞
∫ ∫
∫∫
xx
dxxxdxx
x
dxx
xdxxf
therefore series converges. The same conclusion can be obtained
observing that ∑ n1 and ∑ na has the same nature ),0(1
1∞∈→
n
an .
(b) The general term can be written:
)(1 2 nf
nnan =+
= , where [ ] IR→∞+
= ,1:,1
)( 2 fx
xxf ,
and ∞=+=+
= ∞∞
∞
∫∫ 12
21
1|)1ln(
21
12
21)( xdx
xxdxxf ,
hence ∑∞
= +121n n
n diverges. Besides: ),,0(111 2
2
∞∈→+
=n
n
n
an and after
comparison test, the series diverges.
(c) We have: )(ln1
2 nfnn
an == , where
[ ]xx
xff 2ln1)(,),0(,2: =∞→∞ , and:
71
;2ln
1|ln1
ln)(ln
ln 22 22 2 ∞<=−== ∞∞∞
∫∫ xxxd
xxdx
consequently converges. The same conclusion can be
obtained applying Cauchy’s condensation criterion: since
∑∞
=2nna
)(0 ∞→→ nan the series: and ∑∞
=2nna
∑ ∑∞
=
∞
=
∞
=
∞<==2 1
22
22
12ln
1)2(ln2
122n nn
nnnn
na ∑ (generalized harmonic
series, 12 >=α ) have the same nature.
Exercise 42. Test the series for convergence using D’Alembert’s test:
(a) ...276
94
32
+++ (b) ...531321
31211 +
⋅⋅⋅⋅
+⋅⋅
+
(c) ...642
!542!3
21
+⋅⋅
+⋅
+
Solutions. (a) We have the general term *,32 IN∈= nna nn ,
and
131
31
23
3)1(2
11 <→
+=⋅
+= +
+
nn
nn
aa n
nn
n ,
hence the series ∑∞
=1 32
nn
n is convergent.
(b) !)!12(
!−
=n
nan , and 122121 →
++
=+
nn
aa
n
n , hence
D’Alambert’s test is inconclusive.
But 121
12)1
1222()1(
1
<→+
=−++
=−+ n
nnnn
aa
nn
n and, according to
Raabe-Duhamel’s test, the series diverges.
Exercise 43. By comparing with a harmonic series or a decreasing progression, test the series for convergence:
72
(a) ...54
43
32
21 333
++++ (b) ...4ln
13ln
12ln
1+++
(c) ...531
5211 2 +⋅
+⋅
+
Solutions. (a) The general term is
11 have 1bfor , and ,1 6
16n
3
→==+
=n
bawe
nnna
n
nn ;
But is the general term of harmonic series with n b
161<=α and the series diverges; therefore the given series
diverges too.
∑∞
=1nnb
(b) Using the well known inequality ,ln xx < for all 1>x
we obtain:
2,,1ln1
≥∈∀>= nnnn
an IN ,
hence, by comparison test, the series ∑∞
=2 ln1
n n diverges.
(c) Here the general term is *1 ,
51 IN∈⋅
= − nn
a nn ; but:
151−≤ nna and the geometric series ∑ −15
1n converges
( )1,1(51
−∈=q ), hence ∑∞
=−
115
1n
nn is a convergent series.
Exercise 44. Test the series and ∑ for ∑∞
=1nna
∞
=1
2
nna
convergence where:
(a) 0,,,))...(2)(1(
!≥∈
+++= xxaa
nxxxna n
n IR .
(b) nna n
nln)1(−= .
73
(c) IR∈⋅−⋅⋅⋅−⋅⋅⋅
= aanna n
n ,)14(...73)34(...51 .
Solutions. (a) For the both series converge. Let .
Then: 0=a 0≠a
222
2
2
211
)1()1(,
11 aa
nxn
aa
aanx
na
a
n
n
n
n →++
+=→
+++
= ++
hence, by D’Alambert' test, the both series converge when and diverge for )1,1(−∈a ),1()1,( ∞∪−−∞∈a .
Let 1+=a . Then: xnnx
nnxn
aann
n →+
=⎟⎠⎞
⎜⎝⎛ −
+++
=⎟⎟⎠
⎞⎜⎜⎝
⎛
+ 11
11
1
,
hence by Raabe-Duhamel’s test, ∑ na converges for , and
diverges for
1>x
)1,0[∈x ; if nn a
nax ⋅
+==
11,1 , hence if 1=a the series
diverges (harmonic series) and for 1−=a converges (Leibniz’ criterion). Analogously for ∑> 2,1 nax converges, and if
diverges. If
)1,0[∈x
22
)1(1,1+
==n
ax n and ∑ 2na converges (generalized
harmonic series).
(b) Let IR→∞= ),1[:,ln)( fxxxf . The function is a
decreasing one on [ ]∞,e , since ),(,0ln1)(' 2 ∞∈<−
= exifx
xxf ;
therefore ⎟⎠⎞
⎜⎝⎛
nnln is a decreasing sequence and, by Leibniz’ criterion,
∑∞
=
−1
ln)1(n
n
nn converges. We remark that: 2,,1ln
>∈> nnnn
n IN and
∑∞
=1
1n n
diverges, hence by comparison test it follows that ∑∞
=
−1
ln)1(n
n
nn
is semiconvergent (the series is not absolutely convergent). Since:
74
0ln1
2
23
2
→=nn
n
an , when ∞→n , and ∑∞
=1
2
23
1n
nan
, converges,
we conclude, by comparison test, that the series converges. ∑∞
=1
2
nna
(c) Of course the both series are convergent for . Let
. Then: 0=a
0≠a
aann
aa
n
n →++
=+
34141 , and aa
nn
aa
n
n →++
=+ 22
2
2
21
)34()14( ;
D’ Alambert’s criterion implies that the series converge for and diverges for
)1,1(−∈a),1()1,( ∞∪−−∞∈a .
If }1,1{−∈a since
11816
8161)14()34(1 2
2
2
2
21
2
<++
+=⎟⎟
⎠
⎞⎜⎜⎝
⎛−
++
=⎟⎟⎠
⎞⎜⎜⎝
⎛−
+ nnnn
nnn
aa
nn
n , Raabe-Duhamel’s
test implies that diverges. For ∑∞
=1
2
nna
121
1421
14341,1
1
<→+
=⎟⎠⎞
⎜⎝⎛ −
++
=⎟⎟⎠
⎞⎜⎜⎝
⎛−=
+ nn
nnn
aa
nan
n , hence
diverges, and for
∑∞
=1nna
1−=a , since 11 <+
n
n
aa , for all , Leibniz’
criterion implies the convergence of the series
*IN∈n
∑∞
=1nna
Exercise 45. Test the series with the general term
IR∈−
= ∑=
ain
naun
in ,
ln1
1sin
2, (n>1) for convergence.
Solution. If 00,}|{ ==∈∈ ∑ nn uuZkba and π .
75
Let ∑=−
=∈∉n
in in
Zkka2 ln
11
1,}|{ απ and naan sin= . We verify
the conditions of Abel’s test. First, by Stolz-Cesaro’s criterion we have:
0)1(
)1ln(1
limlim =−−+
=∞→∞→ nn
nnnn
α ; moreover:
,)1ln(
1...3ln
12ln
1)1ln(
ln1...
2ln1
)1ln(1...
2ln1)1(
ln1...
2ln1
11
)1ln(1...
3ln1
2ln11
1
+++<
+⇔
⇔⎟⎠⎞
⎜⎝⎛ ++<⎟⎟
⎠
⎞⎜⎜⎝
⎛+
++−⇔
⇔⎟⎠⎞
⎜⎝⎛ ++
−=<⎟⎟
⎠
⎞⎜⎜⎝
⎛+
+++=+
nnn
nn
nn
nnnn nn αα
and the last inequality is true, since, },...,3,2{)1ln(ln nkfornk ∈+< ; therefore: )( nα is a decreasing sequence, and 0→nα . (1)
Moreover:
2sin
1
2sin
2)1(sin
2sin
sin...2sinsin1 aa
anna
naaaan
kk ≤
+
=+++=∑=
(2)
hence, from (1), (2) and Abel’s test we conclude that the series converges for all ∑∑ = nnn au α IR∈a .
Exercise 46. Study the nature of series:
(a) ∑∞
=13 4cos
n
nπ (b) ( )( ) IR∈⎥
⎦
⎤⎢⎣
⎡ −∑∞
=
xn
nn
,!!2
!!121
α
(c)∑∞
=
∞∈+1
),0[,1n
n xxn
Solutions. (a) Let 4
cos πnan = . Since:
76
18 →na and , when 024 →+na ∞→n it follows that , and, from divergence criterion, the series diverges.
0→/na
(b) The general term of Wallis sequence (see exercise
1.) verifies: nw
212
1!)!12(
!)!2(22
122
ππ≤
+⋅⎥
⎦
⎤⎢⎣
⎡−
=≤⋅+
nnnw
nn
n ,
whence:
nnn bnn
nna
nc :
)12(222
!)!2(!)!12(
)12(2:
2
2
2=⎥
⎦
⎤⎢⎣
⎡++
⋅≤⎥⎦
⎤⎢⎣
⎡ −=≤⎥
⎦
⎤⎢⎣
⎡+
=
ααα
ππ (1)
But:
22
2
2
2)12()1(4lim1lim
αα
αππ⎟⎠⎞
⎜⎝⎛=⎥
⎦
⎤⎢⎣
⎡++
=∞→∞→ n
nn
n
bn
n
n (2)
and: 22
2
1)12(
2lim1
limαα
αππ⎟⎠⎞
⎜⎝⎛=
+=
∞→∞→ nn
n
cn
n
n. (3)
From (1), (2), (3) and comparison test it follows that series , and ∑ nb ∑ nc ∑ na have the same nature. Consequently our
series is convergent if and only if 2>α .
8. EXERCISES
Exercise 1. Prove that:
(a) converges and has the sum ∑∞
=
−⋅1
1
n
nan 2)1(1a−
for
and diverges if . 1|| <a 1|| >a
(b) 1!)!1(
11
=+−∑
∞
=n nn
77
(c) 21
!)!12(1
=+∑
∞
=n nn
(d) 1)!1(1
=−∑
∞
=n nn
(e) ∑∞
=
=+−
1 21
!)!22(!)!12(
n nn
(f) ∑∞
=
>−−−
+=
++++++
11
11
))...(2)(1())...(2)(1(
nabfor
aba
nbbbnaaa
(g) enne
nn
nn5
!,22
! 1
3
1
2
=+= ∑∑∞
=
∞
=
(it is known that enn=∑
∞
=0 !1 ).
Exercise 2. Check whether the necessary condition for
convergence is fulfilled:
(a) ...43
32
21
+++
(b) ...65
43
21
+++
(c) ...76
54
32
+++
(d) ...71
51
31
+++
(e) ...276
94
32
+++
Exercise 3. Test the series for convergence by means of the
integral test:
(a) ...91
511 +++
(b) ...7
14
11 +++
(c) ...43
32
21
333 +++
(d) ...31
121
111
1222 +
++
++
+
78
(e) ...31
321
211
1222 +
++
++
+
(f) ...16
114
112
1222 +−
+−
+−
(g) ...4ln4
13ln3
12ln2
1222 +++
Exercise 4. Test the series for convergence using
d’Alambert’s test:
(a) ...276
94
32
+++
(b) ...!4
8!3
4!2
21 ++++
(c) ...72
352
332
31 3
3
2
2
+⋅
+⋅
+⋅
+
(d) ...8642
!7642
!542!3
21
+⋅⋅⋅
+⋅⋅
+⋅
+
(e) ...34
1333
932
53
1432+
⋅+
⋅+
⋅+
Exercise 5. Using criteria of comparison test the series for
convergence:
(a) ...3
12
11 +++
(b) ...531
5211 2 +⋅
+⋅
+
(c) ...4ln
13ln
12ln
1++
(d) ...1
11
11
1642 +
++
++
+ aaa
(e) ...321
121
113
+++
++
+
(f) ...33
322
25 33 2
7
5 33 2
7
++
++
79
Exercise 6. Find the sum of the series (if it converges):
(a) ...43
132
121
1+
⋅+
⋅+
⋅
(b) ...1071
741
411
+⋅
+⋅
+⋅
(c) ...73
152
131
1+
⋅+
⋅+
⋅
(d) ...!!10!!7
!!8!!5
!!6!!3
!!4!!1
++++
(e) ...)45)(45(
20)45)(45(
20)45)(45(
203344
3
2233
2
22 +−−
+−−
+−−
(f) ...9342
69342
69342
633
3
22
2
+⋅+⋅
+⋅+⋅
+⋅+⋅
arctgarctgarctg
Exercise 7. Test the series with the general term for na
convergence:
(a) 2
2)!( 2
nnna = ; (b) 0, ≥= a
naa
n
n ; (c) 2
)1
( nn
nna+
= ;
(d) nn nna
2)!()!12(
⋅−
= ; (e) ∏= −
−=
n
kn k
ka1 14
34 ; (f) nn
anln
113 15 ⋅
= ;
(g) IR,,ln1
∈= ann
a an αα ; (h) n
na nn
)1ln()1(2 +
−= ;
(i) !!2
!)!12()1(n
na nn
−−= ; (j) IR,sin
ln1
∈= xnxn
an ;
(k) 1
cos2 +
=n
nxan ; (l) ; 0,ln >= aaa nn
(m) , nbn aa = 0,1...
211 >+++= a
nbn ;
(n) ; (o) 0,! >⋅= anaa nn 0,)
1( 2
2
>+
= an
naa nnn ;
(p) ;0,)1
(2
>+
= an
naa nnn (r) ;0,,
!))...(1(
>⋅
++= ba
nnnaaaa bn
80
(s) IR,ln
sincos∈
⋅= x
xnxnx
an ; (t) nnna
310
= ;
(u) nn
n naa 2
1)1(−= ; (v) IR;,,)!)!2(
!)!12(( ∈⋅−
= bann
na ban
(x) IR,,1
∈++
= bann
na b
a
n
Exercise 8. Evaluate the error if we replace the sum of series
with , where , .432 ,, sss ∑=
=n
kcn as
1
*IΝ∈n
(a) ∑∞
=1 !1
n n ; (b) ∑
∞
=
−−1
1 1)1(n
nn
n; (c) ∑
∞
=
−1 !
1)1(n
n
n;
(d) ∑∞
=−
1222n
n
n ; (e) ∑∞
= −1 !)!12(1
n n; ∑
∞
=1 !)!2(1
n n.
Exercise 9. Find a minimal IΝ∈n such the error in the approximation nss ≈ is less than , being the sum, respectively the n
310−nss,
th partial sum of series given at the Exercise 8. Exercise 10. Study the semi convergence of series:
(a) IR,cos)1(13
2 ∈−∑∞
=
xn
xn
n ; (b) ∈−∑∞
=
xn
xn
n ,cos)1(131
IR
(c) nn
nn n2
)1(1
2
2
⋅−∑∞
=
+
; (d) ∑∞
= ++
−11
)1413()1(
n
nn
nn ; (e) ∑
∞
=
−−
15
1)1(n
n
n
(f) ∑∞
=
∈−13
IR,sin)1(n
n xnx ; (g) ∑
∞
=
+−
0 313)1(
nn
n n
(h) ∑∞
=
−
−⋅⋅⋅⋅+
−1
1
)13(...852!)!12()1(
n
n
nn ; (i) ]
2,0[,
)1ln(cos)1(
1
1 π∈
+−∑
∞
=
−
xn
nxn
n
.
Exercise 11. Test the following series for convergence: (a) nn −∑ 1)(ln
(b) ∑ −1)lnlnln( nnn
81
(c) ∑ n2sin π
(d) ∑ )4
tan(nπ
(e) ∑ +−−++ ]11[1 22 nnnnn
(f) ))((ln 45∑ −nn . Exercise 12*. Suppose ∑ na is a convergent series of
positive terms. Prove that ∑ +⋅ 1nn aa converges.
Exercise 13*. ∑∞
=
−+ +−0
211 )1()1(n
n n is a convergent series; form
the square. Does this latter series converge? Do the same for the
series . You should get: ∑∞
=
−+ +−0
11 )1()1(n
n n
∑∞
=
−+ ++++−⋅1
11 ).1...211()1()1(2
n
n
nn
Why does this series converge or not converge?
Exercise 14*. For what values of ∈x IR does the series
∑∞
=
⋅+++1
sin)1...211(1
nnx
nn
converge?
Exercise 15*. If )1,1(−∈r and ∈p IN*, show that the series:
n
pnrpnnnn ⋅−−−∑
∞
+=
))...(2)(1(1
converges.
Exercise 16*. Show that the series with the general term:
)2(
)2(ln1 2
++
=nn
nn
an
converges.
82
Hint. Show that:
∑+= +
<++++
+<
n
mkk mnm
nmm
a1
2ln1
1)2)(1()1)(2(ln
11
Exercise 17*. ∑∞
=
−− +−0
211 )1()1(n
n n is convergent. Show that:
...4
17
15
12
13
11 +−++−+
diverges to infinity.
Exercise 18. Show that the product is divergent
and equal to
∏∞
=
+1
)1(n
n
∞+ .
Exercise 19. Show that ∏∞
=
=−
+2
.1))1(1(n
n
n
Exercise 20. Show that ∏∞
=
+∞=+1
.)11(n n
Exercise 21. .21
)1()2(
12 =
++∏
∞
=n nnn
Exercise 22. ∫ =+
2
0 41
π
πxtg
dxn .
Exercise 23. Show that ∫1
0
]1[ dxx
and ∫ +
1
0 ][dx
xxdx are divergent
generalized integrals.
Exercise 24. Show that converges, then estimate
its value.
∫1
0
][ln dxx
83
Exercise 25. ∫−
=+−1
1 11 πdx
xx
Exercise 26. Show that ∫∞
+`02 )1(
dxxxxarctn converges.
Exercise 27. ∫∞
=0
22 .
34)1,min( dx
xx
Exercise 28. .41
][][3][2132∫
∞
=++ xxx
dx
Exercise 29. ∫∞
+=
++02222 )(2))(( baabbxax
dx π , for all ,
.
IR∈ba,
ba ≠
Exercise 30. ∫ =+
π
π2
044 .22
cossin xdx
Exercise 31. .4
210
4
2
∫∞
=+
πx
dxx
Exercise 32. Show that:
(a) ΖIxx
xxn
n -IR∈=∏∞
=1
,sin2
cos
(b) )11(1∏∞
=
+n nα and ∏
∞
=
−1
)11(n nα IR)∈α( are convergent iff 1. >α
(c) ∏∞
=
=+++
1
4)4(
)3)(1(n nn
nn
(d) ∏∞
=
=−1
2 21)11(
n n
(e) ∏∞
=
=+0
2 2)211(
nn
84
(f) ∏∞
=
=+−
13
3
32
11
n nn
(g)* ∏∞
=
=+1
.02n n
n (Hint ∏=
∈<+
n
k
nnk
k1
,2
12
IN*).
85
