Question

Determine whether the series is absolutely convergent, conditionally convergent
or divergent. Give comprehensive
reasons for your answer, using
appropriate tests (see question for series)

Answer 1

\[\sum_{n=1}^{\infty}(-1)^{n}\frac{ n }{5^{n}(n ^{2}+3)}\]

Answer 2

i found it to be divergent using ratio test fo absolute divergence or convergence but i am doubting my answ

Answer 3

\[\sum_{0}^{\infty} \left| (-1)^n \frac{n}{5^n (n^2+3)} \right| \le \sum_{0}^{\infty} \frac{1}{5^n} \le \sum_{0}^{\infty} \frac{1}{2^n} \le \infty\]Absolutely convergent by comparison test.

Answer 4

damn!!! can u pliz take it step by step

Answer 5

which func are you comparing the series to?

Answer 6

I thought I did.

If we take the absolute value, the (-1)^n term disappears. The fraction n/(n^2+3) is always less than one, so that means the sum of the absolute values have to be less than the sum of the 1/5^n terms.

1/5^n is always smaller than 1/2^n, which sums to 2, less than infinity implies absolutely convergent.

How was that?

Answer 7

and what is the sum?

Answer 8

That wasn't part of the question.

Answer 9

basically the function you are comparing it to is 1/5^n

Answer 10

yeah, and 1/2^n

Answer 11

and is it safe to say 1/5^n is a goemetric series with r=1/5 which is less than 1 hence convergent

Answer 12

Yep.

Answer 13

but i have learnt that you consider the highest power in the numerator and denominator in finding your function to compare

Answer 14

it doesn't seem right to spilt it like that

Answer 15

The fraction n/(n^2+3) is always less than one, so we discarded it on the other side of the inequality sign.

Answer 16

We have the sum of a bunch of terms. Each term has a few factors.

When we take the absolute value, the (-1)^n factor is always one.

The n/(n^2+3) factor is always less than one.

Both of these are multiplied by the 1/5^n factor.

That means the sum has to be less than the sum of the 1/5^n term alone.

Answer 17

Thx i will go thru this stuff again till i get it

Answer 18

Hang tough. Do math every day.

Answer 19

i have had a look an instead of the comparison test i have rechecked my working and come up with convergence here it is below

\[consider \lim_{n \rightarrow \infty}\left| (-1)^{n}\frac{ n }{5^{n}(n ^{2}+3} \right|\]
=\[\frac{ 1 }{ 5 } \lim_{n \rightarrow \infty}\left| \frac{ n ^{3}+3n+n ^{2}+3 }{ n ^{3}+2n ^{2}+4n } \right|\]

which becomes 1/5<1 hence convergent.L

Answer 20

i skipped a few steps. does that look rite?

Answer 21

No, recheck your work. The limit is zero.

Answer 22

Here is my full working check the photo

Answer 23

sorry, this is not my strong point

Answer 24

thx,

Answer 25

your work on limits seems correct to me, L=1/5.

Answer 26

Thx @hartnn, this prob has given me a headache

Answer 27

your welcome ^_^

Answer 28

\[ \lim_{n \rightarrow \infty} \frac{n}{5^n(n^2+3)} \neq 1/5\]
if you meant you used this term in the ratio test, where you divided the (n+1)th term by the nth term, then that would yield 1/5.

Answer 29

You came in 3/4 way into a discussion follow the thread and you will notice that i was talking about using ratio test fo absolute divergence or convergence

Answer 30

That's fine, but looking at your post you said
\[ \lim_{n \rightarrow \infty} \left| (-1)^n \frac{n}{5^n(n^2+3)}\right| =\lim_{n \rightarrow \infty} (\text{blah blah blah}) = 1/5\]
Which is not true, because the limit on the left is zero. Forgive my confusion, glad you got it worked out.