Question

Okay. I'm having trouble with series. Here's the problem (I'm looking to see whether the series is convergent or divergent, if convergent I have to find the sum it converges to): summation from n=1 to infinity of \(\ \frac{n}{n+5} \).

Answer 1

I'm thinking partial fractions or factoring n out but I'm not sure...

Answer 2

can you type this out i can't read it. Sorry :(

Answer 3

?

Answer 4

nvm, i got it.
You can view this as a geometric series.\[n/(n-(-5))\]

Answer 5

so your first term is n and your ratio is -5. now using the formula you can find the sum it converges to.

Answer 6

So you factor the n out?

Answer 7

???

Answer 8

not really. are you familiar with the formula to find the sum of a convergent series ?

Answer 9

I meant to find what a and r is before I find the sum.

Answer 10

Because r is less than 1 (-5) does that mean this is divergent?

Answer 11

What is the limit of the TERMS as n increases without bound? If it's not zero, it doesn't converge.

Answer 12

But r has to be -1<r<1 right?

Answer 13

r is -5 so...

Answer 14

Yes, that is good for 'r', but you don' even need that. Just look at the limit of the terms. Test #1, in ALL cases.

Answer 15

BTW, r is NOT 5. It's 1.

Answer 16

.
? And wouldn't the limit as n approaches infinity be 0 because denominator is larger?

Answer 17

No. The limit of the terms is 1. As n increases, the 5 becomes insignificant.

Answer 18

Okay. And how do I find r to be one then?

Answer 19

the coefficient of numerator and the denominator is 1
therefore, the limit is one.

Answer 20

try graphing it.

Answer 21

First, I don't actually care. Since the limit of the terms is 1, it cannot converge. There is no reason to do the ration test.

Second, just do it. \(\dfrac{\dfrac{n+1}{n+6}}{\dfrac{n}{n+5}} = \dfrac{n^{2}+6n+5}{n^{2}+6n}\). The limit of this, as n increases without bound, is 1. As n increases, 5 becomes insignificant and it is clearly 1.