Question

Convergence/divergence

Answer 1

I don't get what the question is asking me for the limit section.

Answer 2

I think it's supposed to be hinting at what convergence test you should be using.

Answer 3

Probably the limit comparison test.

Answer 4

I know but I don't get what it's asking me to type in. It doesn't make sense.

Answer 5

In the box? I would expect a series you might compare the given series to.

For example, if you had been given
\[\frac{1}{n^2+14n}\]
you'd probably compare it to
\[\frac{1}{n^2}\]

Answer 6

What's with the fancy numerator then?

Answer 7

If you're familiar with the LCT, then you should know that given some series
\[\sum a_n,\]
choose some series
\[\sum b_n\]
and check to see if the limit is a positive, finite number:
\[\lim_{n\to\infty}\frac{a_n}{b_n}=L\in(0,\infty)\]
The numerator is just the given series.

Answer 8

In my example, you would check
\[\lim_{n\to\infty}\frac{\frac{1}{n^2+14n}}{\frac{1}{n^2}}=\lim_{n\to\infty}\frac{n^2}{n^2+14n}=1\]
Thus, my example series converges.

Answer 9

Ahh I see.

Answer 10

Try comparing to
\[\sum_{n=1}^\infty\frac{1}{n^\frac{7}{3}}\]

Answer 11

So I guess I would use \[\frac{ 8 }{ n }\] as a comparison?

Answer 12

Why that series?

Answer 13

The trick to picking b_n has to do with the highest degree term of the polynomial. In this case, it's 10/3. (Since you have n^10 to a power of 1/3)

Dividing by the highest power in the numerator, the degree of the denominator is 10/3 - 1 = 7/3.

Answer 14

Well for large values of n I can treat the series as 8/n .

Answer 15

then use 8/n, and learn from your own experience.

Answer 16

I only have 1 shot to get it right :( .

Answer 17

I don't want to be wrong.

Answer 18

in that case, learn from the more experienced, and try 1/n^(7/3) as suggested by @SithsAndGiggles

Answer 19

@SithsAndGiggles @sirm3d : Whoops I was looking at the wrong question when I said my answer hehe...

Answer 20

Ahh I think I get it...