Question

\[\sum_{k=1}^{\infty}3*\frac{ 4^k }{ 7^k }\]

Answer 1

You have a geometric series of the form \(\sum ar^n\), which converges to \(\dfrac{a}{1-r}\) for \(|r|<1\).

Answer 2

That's slightly confusing. I am asked to "Evaluate" this. I'm not sure what I'm supposed to write there.

Answer 3

\[\sum_{n=0}^\infty ar^n=a+ar+ar^2+\cdots=\frac{a}{1-r}\]
for \(|r|<1\). In this case, \(a=3\) and \(r=\dfrac{4}{7}\), which is between -1 and 1, and so the series converges.

Note however that the series I wrote starts at 0, whereas yours starts at 1. Make a minor adjustment:
\[\sum_{\color{red}{n=1}}^\infty ar^n=ar+ar^2+\cdots=\frac{a}{1-r}\color{red}{-a}\]

Answer 4

so as long as it's over 0, it will always converge? It's that simple?

The next question goes like this: Determine whether the series converges or diverges:

\[\sum_{k=1}^{\infty}\frac{ 1+3^k }{ 2^k }\]

Answer 5

use the ratio test, find limit as n approaches infinity of $$\frac{a_{n+1}}{a_{n}}$$

Answer 6

replace n with k in your case

Answer 7

I don't understand ratio, divergence or any other tests at all. My teacher is not helpful at all and the online explanations suck.

Answer 8

ok so do you know how to find $$a_{k+1}$$ and $$a_{k}$$?

Answer 9

Not really :(