# Infinite series problem

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.

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

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.

HallsofIvy
Homework Helper
You say "thanks for the replies" but simply ignore them?

$$\frac{7^{k+1+1}}{2^{3k-1+1}}$$
$$\frac{7^{(k+1)+1}}{2^{3(k+1)-1}}= \frac{7^{k+2}}{2^{3k+2}}$$.