- #1

- 18

- 0

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.

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- Thread starter CalculusSandwich
- Start date

- #1

- 18

- 0

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.

- #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

- 18

- 0

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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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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