Finding the general term of the series

  • #1
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Homework Statement



Find the general term and test the nature(convergent/divergent) of:
1/3 + 2/15 + 2/35 + .....

The Attempt at a Solution


If I simplify I get ,
1/3 + (1)(2)/(3)(5) + (1)(2)(3)/(3)(5)(7) +.............

I found that,
TERM(N) = TERM(N-1)((N)/(2N+1)

After this I am struck.
 

Answers and Replies

  • #2
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The new format of this website is difficult for me.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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You see, I suspect, that the numerator of the nth term is n! The denominator is something like a factorial except that it involves only odd integers- okay, make it a factorial by inserting the even integers:
[tex]\frac{1}{3*5*7}= \frac{2*4*6}{2*3*4*5*6*7}= \frac{2*4*6}{7!}[/tex]

Now, factor a 2 out of each term in the numerator: 2*4*6= 2(1)*2(2)*2*3= 2^3(3!).

So [tex]\frac{1}{3*5*7}= \frac{2*4*6}{2*3*4*5*6*7}= \frac{2^3(3!)}{7!}[/tex]

Your term [tex]\frac{1*2*3}{3*5*7}= \frac{2^3(3!)^2}{7!}[/tex].

Can you write the nth term now?

Do you see n
 
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  • #4
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Can you write the nth term now?

Do you see n

yeah... i can write the nth term and i found that the series is convergent.

But this series:
1/4 + (1)(5)/(4)(8) + (1)(5)(9)/(4)(8)(12) + ...

The general term for the denominator is 4^n (n!)

but the numerator terms are tricky : 1*5*9

==> 1*2*3*4*5*6*7*8*9/2*3*4*6*7*8

==>9!/2*3*4*6*7*8

how do i simplify this part : 2*3*4*6*7*8 ?
 

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