Question

Does this series converge BY THE (LIMIT) COMPARISON TEST?:
http://www.tiikoni.com/tis/view/?id=4f80899

Answer 1

@satellite73: This is the other thread.

Answer 2

Are both the LIMIT and NON-LIMIT comparison tests not applicable here?

Answer 3

Using b_n = 1/n, I get lim n-> inf = 0 and I'm supposed to get 0 < c < inf according to Wikipedia.

Answer 4

Diverges by the comparison test, as \((\ln(n))^5>1\)

Answer 5

i.e. you can compare it to \(\sum\frac{1}{n}\)

Answer 6

How can I formally ignore the +9?

Answer 7

the \(+9\) is unimportant

Answer 8

compare for large ns:

100000000000000009 and 100000000000000000

Answer 9

So can I just compare the series in the image to 1/(n+9) instead of 1/n?

Answer 10

if you like

Answer 11

Am I not supposed to show that 1/(n+9) diverges though?

Answer 12

\[\frac{1}{10000000000000009}\approx\frac{1}{10000000000000000}\]

Answer 13

(in order to use it)

Answer 14

Amistre64, I get the intuition, but I wouldn't want to lose a mark from my teacher, for example, for comparing to 1/(n+9) instead of 1/n.

Answer 15

Unless you guys think that, it doesn't need to be shown in order to be used.

Answer 16

Oh wait!

Answer 17

n+9 is a right shift .... it has no effect on the overall value for large values of n

Answer 18

I'll just use the limit comparison test with 1/(n+9) and 1/n ... problem solved!

Answer 19

Just to clarify, I meant, now I have shown that 1/(n+9) diverges.

Answer 20

... so I can use it to compare to the image!