Sequence

Replies

• 

  Saandeep Sreerambatla

  1+2 + 2+ 1+ 0.333 +0.0833 + 0.01677 + 0.0027
  
  May be solving for few more steps we get a value almost equal to zero as the next term.
  
  
  Is this the answer expecting?
• 

  speedx

  thnks ya .........but if u can SHOW by COMPARISON TEST (infinite seriies ) method it wil be better 😁
• 

  mech_guy

  Hi speedx,
  
  

  speedx

  TEST THE CONVERGENCE OF THE SERIES
  1+ 2/1! + 2^2/2! + 2^3/3! +.............
  
  show by steps 😒

  Series is: A(n) = 2^n / (n!)
  
  To prove its convergence, RATIO TEST will be best which states:
  
  

  If SIGMA(A(n)) be a series with positive terms and
   lim(n->inf) A(n+1)/A(n) = L
  Then,
  If L<1, then SIGMA(A(n)) converges
  If L>1, then SIGMA(A(n)) diverges
  If L=1, then SIGMA(A(n)) may converge or diverge.

  So in this case A(n+1) = 2^(n+1)/(n+1)!
  A(n) = 2^n/n!
  A(n+1)/A(n) = 2/n -> 0 for lim(n->inf)
  
  Therefore, given series converges for all n.
  
  

  but if u can SHOW by COMPARISON TEST (infinite seriies ) method it wil be better

  Well Comparison Test too can be used here but i think RATIO TEST will be more convenient in this case:-
  
  COMPARISON TEST says:
  

  Consider two series A(n) and B(n):-
  If for very large values of "n", 
  If B(n) >= A(n) then If B(n) converges, A(n) too will converge
  If A(n) <= B(n) then If A(n) diverges, B(n) too will diverge.
  
  That is, if larger of the two series "converges" so will the smaller series.
  If smaller series "diverges" so will the larger series.

  Now considering the present series at hand, A(n) = 2^n / n!
  
  We know that for n = 4, "n!" will overtake "2^n", so we can make another series,
  B(n) = 2^4 / n
  We have made this B(n) so that B(n) >= A(n) for very large values of "n".
  
  Now we know by LIMIT Rule (n-> inf) that 2^4/n converges.
  Therefore A(n) also converges.
• 

  Anil Jain

  mech_guy... you are a mathematics genius. Hats off to you.
  
  -CB
• 

  mech_guy

  Hi CB,
  
  

  crazyboy

  mech_guy... you are a mathematics genius. Hats off to you.
  
  -CB

  Oops, that was a bit too much 😀
  
  We all studied this interesting topic of "Convergence of Series" in one of our sem courses. I just googled to revisit what we studied and everything was simple then.
  
  Thanks to "speedx" for posting such question, it made me "keep in touch" with basics 😀
  
  Cheers