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## Study23 2 years ago 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 $$\ \frac{n}{n+5}$$.

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1. Study23

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

2. macknojia

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

3. Study23

?

4. macknojia

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

5. macknojia

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

6. Study23

So you factor the n out?

7. Study23

???

8. macknojia

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

9. Study23

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

10. Study23

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

11. tkhunny

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

12. Study23

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

13. Study23

r is -5 so...

14. tkhunny

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.

15. tkhunny

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

16. Study23

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

17. tkhunny

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

18. Study23

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

19. macknojia

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

20. macknojia

try graphing it.

21. tkhunny

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.

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