2017-2018吉林大学数学分析第一学期期中考试参考答案

k=1
1 , n = 1, 2, · · · . K {xn } uÑ. k
n+p
é ?‰
ê p,
0
|xn+p − xn | =
k=n+1
1 k
p → 0, n+1
n → ∞.
4.
{xn }, {yn } ´ü‡¢ê
, {xn } î‚üN4Oªu +∞. e lim
n→∞
xn+1 − xn xn Ø•3, K lim n→∞ yn yn+1 − yn
oK$Ž, k a3 = 2 + a + a2 , )
a = 2, = lim xn = 2.
n→∞
x1 = 2 ž, xn ≡ 2 (n = 1, 2, · · · ), dž lim xn = 2. x1 > 2 ž. ·‚äó 2 2 xn =
3
xn
x1 (n = 1, 2, · · · ), ¯¢þ, e x1
(xn − 2)(x2 n + xn + 1)
3
2 (2 + xn + x2 n ) + xn
3
2 2 + xn + x2 n + xn
0,
= xn+1 ê 4•
(2) (3)
xn . ¤± {xn } üN4O. düNk.
n, {xn } Âñ.
n→∞
n→∞
2 lim xn = a. d x3 n+1 = 2 + xn + xn 9
n→∞
n→∞
x1 + x2 + · · · + xn = ∞. n
n→∞
)
†Ø. ~X
lim
xn = (−1)n n, n = 1, 2, · · · , K lim xn = ∞. Ï• x1 + x2 + · · · + x2k+1 −k − 1 1 = lim =− , k→∞ k→∞ 2k + 1 2k + 1 2 lim
n2 + n − πn) = lim sin2 √
n→∞
n2
nπ π = sin2 = 1. 2 +n+n
2. lim
n→∞
1+
1 1 + 2 n n
.
)
Ï•
1+ 1 n
n
1+ 1 n−1
n
1 1 + n n2
n
1+
1 n−1
n
,
n = 1, 2, · · · 1+ 1 1 + n n2
n
… lim
•Ø•3. ) †Ø. ~X
xn = n, yn =
n
[(−1)k + 2],
k=1
n = 1, 2, · · · ,
K
n→∞
lim
xn+1 − xn 1 = lim (Ø•3). n→∞ (−1)n+1 + 2 yn+1 − yn lim x2n+1 2n + 1 1 = lim = , n→∞ 2(2n + 1) − 1 y2n+1 2
,
n > N ž,
a0 n2017 + a1 a0 |a1 b0 − a0 b1 | |a1 b0 − a0 b1 | + 1 − = b0 n2017 + b1 b0 |b0 ||b0 n2017 + b1 | |b0 |(|b0 |n2017 − |b1 |) 2(|a1 b0 − a0 b1 | + 1) 2(|a1 b0 − a0 b1 | + 1) < ε. |b0 |2 n2017 |b0 |2 n
x1 + x2 + · · · + x2k 1 1 = lim = , k→∞ 2 k→∞ 2k 2
¤± lim
x1 + x2 + · · · + xn = ∞. n→∞ n
3. eé?‰
ê p, Ñk lim (xn+p − xn ) = 0, K {xn } Âñ.
n→∞ n
)
†Ø. ~X
xn =
2 xn =
3
xn
3
2 (n = 1, 2, · · · ), ¯¢þ, e x1 2 + x1 + x2 1
3
xn−1
3
2 + xn−1 + x2 n−1
2x1 + x2 1
3
2x2 1
Ùg, d x1
xn
2 (n = 1, 2, · · · ), k
3
xn+1 − xn =
2 + xn + x2 n − xn = −
n→∞
1+
1 n
n
= lim
2
n→∞
1+
= e, ¤±dY@½n, lim
n→∞
= e.
3. lim (n!)1/n .
n→∞
)
kOŽ lim ln[(n!)1/n ], = lim
n→∞
2
n→∞
ln 2 + ln 3 + · · · + ln n . n2 yn = n2 , n = 1, 2, · · · , √
xn = ln 2 + ln 3 + · · · + ln n,
K
n→∞
lim
xn+1 − xn ln(n + 1) = lim = lim n →∞ n→∞ yn+1 − yn 2n + 1
n+1 ln 2n + 1
n+1
n+1
= 0.
Šâ Stolz ½n,
n→∞
lim
ln 2 + ln 3 + · · · + ln n xn = lim = 0. n→∞ yn n2
n→∞
lim
x2n 2n 1 = lim = , n→∞ 4n y2n 2
n→∞
¤± lim
n→∞
xn 1 = . yn 2
1
! OŽe
2 n→∞
4• (zK 10 ©,
n2 + n).
30 ©).
1. lim sin (π
)
n→∞
lim sin2 (π
n2 + n) = lim sin2 (π
n→∞ n
3
2 (2 + xn + x2 n ) + xn
3
2 2 + xn + x2 n + xn
0,
= xn+1
xn . ¤± {xn } üN4~. düNk.
n→∞
n, {xn } Âñ. Ón (1), Œ
n→∞
lim xn = 2.
nþ¤ã, {xn } Âñ… lim xn = 2
3
3
x n−1
3
2, K
3
2 + xn−1 + x2 n−1
2 + x1 + x2 1
3
2x1 + x2 1
2x2 1
x3 1 = x1
Ùg, d 2
xn
x1 (n = 1, 2, · · · ), k
3
xn+1 − xn =
2 + xn + x2 n − xn = −
(xn − 2)(x2 n + xn + 1)
U½Â, lim
a0 n2017 + a1 a0 = . n→∞ b0 n2017 + b1 b0 n π = . n+1 2 ε ∈ 0, π , 2 N= 1 + 1, 1 − cos ε n > N ž, 1 n
2. lim arcsin
n→∞
y²
é?‰
arcsin
n π π n 1 − = − arcsin = arccos 1 − n+1 2 2 n+1 n+1 n π = . n+1 2 2
u´ lim (n!)1/n = 1.
n→∞
2
n! U½Ây²e
1. lim
4• (zK 10 ©,
20 ©).
a0 n2017 + a1 a0 = , Ù¥ a0 b0 = 0. 2017 n→∞ b0 n + b1 b0
y²
é?‰
ε > 0,
N = 1 + max
2(|b1 | + 1) 2(|a1 b0 − a0 b1 | + 1) , |b0 |2 ε |b0 |
xn+1 − xn+2 = (xn + xn+1 ) − (xn + xn+2 ),
百度文库
¤± {xn+1 − xn+2 } Âñ, = {xn − xn+1 } Âñ. q
xn = 1 [(xn + xn+1 ) + (xn − xn+1 )], 2
¤± {xn } Âñ.
2. e lim xn = ∞, K lim
2017–2018 ÆcêÆ©ÛJ™‘§ I
5 ê 4• 6 ÿÁK땉Y
2017 ?/^Ÿ• 2017.11.13
˜! ä¿`²nd (zK 9 ©, 36 ©). 1. e {xn + xn+1 } Ú {xn + xn+2 } ÑÂñ, K {xn } Âñ. ) (. Ï• {xn + xn+1 } Ú {xn + xn+2 } ÑÂñ, …
arccos 1 −
< arccos(cos ε) = ε.
U½Â, lim arcsin
n→∞
o! (14 ©) ) (1)
x1 > 0, xn+1 =
3
2 + xn + x2 n , n = 1, 2, · · · .
äê
{ xn }
ñÑ5.
2, K x3 1 = x1
©n«œ/?Ø. 0 < x1 < 2 ž, ·‚äó x1