# Infinite series problem

1. Nov 19, 2006

### CalculusSandwich

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.

2. Nov 19, 2006

### d_leet

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: Nov 19, 2006
3. Nov 19, 2006

### Office_Shredder

Staff Emeritus
A simpler way, see if you can re-write it as a series in terms of (7/8)k

4. Nov 22, 2006

### CalculusSandwich

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. Nov 22, 2006

### HallsofIvy

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
$$\frac{7^{k+1+1}}{2^{3k-1+1}}$$
is wrong. You are adding 1 to k, not to the exponent. It should be:
$$\frac{7^{(k+1)+1}}{2^{3(k+1)-1}}= \frac{7^{k+2}}{2^{3k+2}}$$.

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!