Page 36 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 2] SEQUENCES 27

LIMIT OF A SEQUENCE
3n 1
. (a) Write the 1st, 5th, 10th, 100th, 1000th,
4n þ 5
2.3. A sequence has its nth term given by u n ¼
10,000th and 100,000th terms of the sequence in decimal form. Make a guess as to the limit of
this sequence as n !1.(b) Using the definition of limit verify that the guess in (a)is actually
correct.
n ¼ 1 n ¼ 5 n ¼ 10 n ¼ 100 n ¼ 1000 n ¼ 10,000 n ¼ 100,000
:22222 ... :56000 ... :64444 ... :73827 .. . :74881 .. . :74988 ... :74998 .. .
ðaÞ
3
A good guess is that the limit is :75000 ... ¼ . Note that it is only for large enough values of n that
4
a possible limit may become apparent.
(b)We must show that for any given > 0(no matter how small) there is a number N (depending on )
3
such that ju n j < for all n > N.
4

3n 1 19 19
3
Now ¼ < when < or
4n þ 5 4 4ð4n þ 5Þ 4ð4n þ 5Þ

1 19 1 19
> ; 4n þ 5 > ; n > 5
4ð4n þ 5Þ
19 4 4 4
3
1
3
Choosing N ¼ ð19=4 5Þ,we see that ju n j < for all n > N,sothat lim ¼ and the proof is
4 4 4
complete. n!1
1
1
Note that if ¼ :001 (for example), N ¼ ð19000=4 5Þ¼ 1186 . This means that all terms of the
4 4
sequence beyond the 1186th term differ from 3 4 in absolute value by less than .001.
c
2.4. Prove that lim ¼ 0 where c 6¼ 0 and p > 0 are constants (independent of n).
n!1 n p
p
We must show that for any > 0there is a number N such that jc=n 0j < for all n > N.
1=p 1=p

< when < , i.e., n > or n > (depending on ), we
c jcj p jcj jcj jcj
Now p p . Choosing N ¼
n n
p
p
see that jc=n j < for all n > N,proving that lim ðc=n Þ¼ 0.
n!1
1 þ 2 10 n 2
2.5. Prove that lim ¼ .
n!1 5 þ 3 10 n 3
n
1 þ 2 10 2
We must show that for any > 0there is a number N such that < for all n > N.
5 þ 3 10 n 3
n
1 þ 2 10 7 7
2 n
Now ¼ < when < , i.e. when 3 ð5 þ 3 10 Þ > 1= ,
n
5 þ 3 10 n 3 3ð5 þ 3 10 Þ 3ð5 þ 3 10 Þ 7
n
n
1
1
n
3 10 > 7=3 5, 10 > ð7=3 5Þ or n > log f ð7=3 5Þg ¼ N,proving the existence of N and thus
8 10 3
establishing the required result.
Note that the above value of N is real only if 7=3 5 > 0, i.e., 0 < < 7=15. If A 7=15, we see that
n
1 þ 2 10
2
< for all n > 0.
5 þ 3 10 3
n
2.6. Explain exactly what is meant by the statements (a) lim 3 2n 1 ¼1,(b) lim ð1 2nÞ¼ 1.
n!1 n!1
(a)Iffor each positive number M we can find a positive number N (depending on M)suchthat a n > M for
all n > N,then we write lim a n ¼1.

1 log M
n!1
In this case, 3 2n 1 > M when ð2n 1Þ log 3 > log M; i.e., n > þ 1 ¼ N.
2 log 3
(b)Iffor each positive number M we can find a positive number N (depending on M)suchthat a n < M
for all n > N,then we write lim ¼ 1.
n!1
1
In this case, 1 2n < M when 2n 1 > M or n > ðM þ 1Þ¼ N.
2