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Consider if the following Infinite Series converges or diverges; If converges, find the sum.
[tex] \displaystyle \large{ 1 + 2 + \frac{1}{2} - \frac{2}{3 } + \frac{1}{4} + \frac{2}{9} + \frac{1}{8} - \frac{2}{27} + ...}[/tex]
Please show your work too. Thanks!​

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kamilmatematik100504

Answer: 3

Step-by-step explanation:

  • You can immediately notice that the sum consists of the sum of two infinitely decreasing geometric progressions
  • [tex]\displaystyle \large \boldsymbol {} \underline1+\underline 2+\underline {\frac{1}{2}} -\frac{2}{3} +\underline {\frac{1}{4}} -\frac{2}{9} +\underline{\frac{1}{8}} -\frac{2}{27} + \ldots = \\\\\\ \underbrace{\bigg(1+2+\frac{1}{2} +\frac{1}{4} \ldots \bigg)}_{\rm \big S_2}-2 \underbrace{\bigg(\frac{1}{3}+\frac{1}{9} +\frac{1}{27} \ldots \bigg)}_{\big S_3}[/tex]  

  • The formula for an infinitely decreasing geometric progression:
  • [tex]\displaystyle \large \boldsymbol { \rm S=\frac{b_1}{1-q} \ \ ; \ \ |q|<1}[/tex]
  • Where :
  • q -denominator
  • b₁ -is the first member of the progression
  • Then:
  • Denote the sum of the numbers in the first bracket for S₂ ; and in the second bracket for S₃
  • [tex]\displaystyle \large \boldsymbol {} \rm S_2=\frac{2}{1-\dfrac{1}{2} } =\frac{2}{\dfrac{1}{2} } =4 \\\\\\\ S_3=\frac{\dfrac{1}{3} }{1-\dfrac{1}{3} } = \dfrac{1}{\dfrac{3}{\dfrac{2}{3} } } =\frac{1}{2}[/tex]
  • Substitute the values :

  • [tex]\large \boldsymbol {} \rm S_2-2S_3=4- 2\cdot \dfrac{1}{2} =\boxed{3}[/tex]

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