Infinite series problem

  • #1
Hello i have the infinite series

7^(K+1)/2^(3k-1)

How do i find what it converges to if it does converge.

Limit comparison does me no good. I am thinking integral and ratio test.

root test does me no good either.
 

Answers and Replies

  • #2
1,074
1
Think geometric series, the ratio test also works nicely and is probably easier than trying to make this look more like a geometric series. The root test should work as well, but I think it would be a little tricky.
 
Last edited:
  • #3
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
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A simpler way, see if you can re-write it as a series in terms of (7/8)k
 
  • #4
Thanks for the replies, I applied the ratio test.

7^K+1+1 2^3k-1 7^k (x) 7^2 2^3k (x) 2^-1
--------- x ---------- = ----------------- x --------------
2^3k-1+1 7^k+1 2^3k 7^k (x) 7^1

-----------
7^k+1
------
2^3k-1


So everything but the 7^2 which is 49 and 2^-1 / 7 which is 24.5 / 7 , which gives me 3.5, however i think this is wrong.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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You say "thanks for the replies" but simply ignore them?

Your calculation is completely wrong:
7^K+1+1 2^3k-1 7^k (x) 7^2 2^3k (x) 2^-1
--------- x ---------- = ----------------- x --------------
2^3k-1+1 7^k+1 2^3k 7^k (x) 7^1
[tex]\frac{7^{k+1+1}}{2^{3k-1+1}}[/tex]
is wrong. You are adding 1 to k, not to the exponent. It should be:
[tex]\frac{7^{(k+1)+1}}{2^{3(k+1)-1}}= \frac{7^{k+2}}{2^{3k+2}}[/tex].

In any case, it is far simpler to do as both d_leet and Office Shredder suggested: write this as a geometric sequence with common ratio 7/8. That way, it is not only obvious that the sequence converges but easy to see what it converges to!
 

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