Question

Determine whether the series converges or diverges

Answer 1

\[\sum_{n=0}^{\infty}\frac{ 5*7^{n+1} }{ 2^{3n-1} }\]

Answer 2

well a simple way is to look at the terms


1st term

35/0.5 = 70

2nd term

245/4 = 61.25

3rd term

1715/32 = 53.938

the common ratio is < 1 so it should converge.

Answer 3

Ahh yes. That's what I was looking for. Thanks!

Answer 4

There's another more algebraic way to look at it if you're interested lol

Answer 5

(I hate computing numbers lol)

Answer 6

Sure. Go for it.

Answer 7

\[\sum_{n=0}^{\infty}\frac{5*7*7^n}{2^{-1}*2^{3n}}\]=\[5*7*2^{-1}\sum_{n=0}^{\infty}\frac{7^n}{2^{3n}}=5*7*2^{-1}\sum_{n=0}^{\infty}(\frac{7}{2^3})^n\]

So you have a geometric series with ratio r = 7/8 and you know that a geomric series converges for |r|<1 and 7/8<1 :)

Answer 8

geometric series*

Answer 9

That's a nice proof :) .

Answer 10

I wish I could give both of you medals XD .

Answer 11

Hehe it's all right :P It seemed long to type lol because of th LaTeX but I find it shorter and I don't have to rely on a calculator (which is useful if your prof doesn't allow it :)... I never was able to use calculators in my cal classes so I always rely on these methods)>.. of course you can divide those numbers by hand, but sometimes the numbers might be trickier

Answer 12

Yeah... we can't use calculators either. But I can calculate by hand really well so I don't mind XD .

Answer 13

One mistake you made.

Answer 14

It should be (5*7)/(2^-1) or 5*7*2.

Answer 15

Oh dear you are absolutely right! :) My mistake I'm sorry. I didn't care too much about the constants though they won't affect convergence anyway ;)

Answer 16

I know :) . Just because I am OCD I thought I should point that out :P .

Answer 17

Yeah you are right to point it out though :) Picky profs would probably take off marks (lol this is exactly how I lose marks sometimes, careless fast calculations =_=)