Question

Does 7+28+112+448+...
converge or diverge?

Answer 1

This is a series I can put into the form of ar^n, where a is the initial value and r^n is basically the pattern of the series. I can fully write the series like this:

\[\sum_{i=0}^{\infty}ar ^{n}\] Where 7 is your a and 4 is your r, because that is the pattern here, multiplying by 4.

\[\sum_{i=0}^{\infty}7(4)^{n}\]
Now when the absolute value of r, 4 in this case, is greater than or equal to 1, your series diverges. So this series diverges. If your absolute value of r was 0<r<1, the series would converge. If your series converges, your sum is equal to:

\[\frac{ a }{ 1-r }\]
Of course it diverges here so you don't need that, but I thought I would put it :P

Answer 2

thank you

Answer 3

Np :3