anonymous
  • anonymous
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 \(\ \frac{n}{n+5} \).
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I'm thinking partial fractions or factoring n out but I'm not sure...
anonymous
  • anonymous
can you type this out i can't read it. Sorry :(
anonymous
  • anonymous
?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
nvm, i got it. You can view this as a geometric series.\[n/(n-(-5))\]
anonymous
  • anonymous
so your first term is n and your ratio is -5. now using the formula you can find the sum it converges to.
anonymous
  • anonymous
So you factor the n out?
anonymous
  • anonymous
???
anonymous
  • anonymous
not really. are you familiar with the formula to find the sum of a convergent series ?
anonymous
  • anonymous
I meant to find what a and r is before I find the sum.
anonymous
  • anonymous
Because r is less than 1 (-5) does that mean this is divergent?
tkhunny
  • tkhunny
What is the limit of the TERMS as n increases without bound? If it's not zero, it doesn't converge.
anonymous
  • anonymous
But r has to be -1
anonymous
  • anonymous
r is -5 so...
tkhunny
  • 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.
tkhunny
  • tkhunny
BTW, r is NOT 5. It's 1.
anonymous
  • anonymous
. ? And wouldn't the limit as n approaches infinity be 0 because denominator is larger?
tkhunny
  • tkhunny
No. The limit of the terms is 1. As n increases, the 5 becomes insignificant.
anonymous
  • anonymous
Okay. And how do I find r to be one then?
anonymous
  • anonymous
the coefficient of numerator and the denominator is 1 therefore, the limit is one.
anonymous
  • anonymous
try graphing it.
tkhunny
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.