Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

4.E: Convergence of Sequences and Series (Exercises)

( \newcommand{\kernel}{\mathrm{null}\,}\)

Q1

Prove that if limnsn=s then limn|sn|=|s|. Prove that the converse is true when s=0, but it is not necessarily true otherwise.

Q2

  1. Let (sn) and (tn) be sequences with sntn,n. Suppose limnsn=s and limntn=t. Prove st. [Hint: Assume for contradiction, that s>t and use the definition of convergence with ε=\(fracst2 to produce an n with sn>tn.]
  2. Prove that if a sequence converges, then its limit is unique. That is, prove that if limnsn=s and limnsn=s, then s=t.

Q3

Prove that if the sequence (sn) is bounded then limn(snn)=0.

Q4

  1. Prove that if x1, then (4.E.1)1+x+x2++xn=1xn+11x
  2. Use (a) to prove that if |x|<1, then limn(j=0nxj)=11x

Q5

Prove (4.E.2)limna0+a1n+a2n2++aknkb0+b1n+b2n2++bknk=akbk

provided bk0. [Notice that since a polynomial only has finitely many roots, then the denominator will be non-zero when n is sufficiently large.]

Q6

Prove that if limnsn=s and limn(sntn)=0, then limntn=s.

Q7

  1. Prove that if limnsn=s and s<t, then there exists a real number N such that if n>N then sn<t.
  2. Prove that if limnsn=s and r<s, then there exists a real number M such that if n>M then r<sn.

Q8

Suppose (sn) is a sequence of positive numbers such that limn(sn+1sn)=L

  1. Prove that if L<1, then limnsn=0. [Hint: Choose R with L<R<1. By the previous problem, N such that if n>N, then sn+1sn<R. Let n0>N be fixed and show sn0+k<Rksn0. Conclude that limksn0+k=0 and let n=n0+k.]
  2. Let c be a positive real number. Prove (4.E.3)limn(cnn!)=0

This page titled 4.E: Convergence of Sequences and Series (Exercises) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Eugene Boman and Robert Rogers (OpenSUNY) via source content that was edited to the style and standards of the LibreTexts platform.

Support Center

How can we help?