Question

Show whether the series converges or diverges using an appropriate series test or tests, Calculus B...

Answer 1

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

Answer 2

alternating series test

Answer 3

You can write your series as:
\(\color{#000000}{ \displaystyle \sum_{ n=1 }^{ \infty }(-1)^{n}\frac{\ln(n)}{n+3}}\)

Answer 4

One note that (I checked) it won't be monotonically decreasing
starting from n=1 (for instance |a2|<|a3|), but since
\(\color{#000000}{ \displaystyle \lim_{n\to \infty}\frac{\ln(n)}{n+3}}\)

therefore there is going to be some n=j starting from which
\(\color{#000000}{ \displaystyle \frac{\ln(n)}{n+3}}\) is monotonically decreasing.

that will show that starting from n=j, your series ___________ ?
and therefore starting from n=0 the series also ________?

Answer 5

The alternating series test would be inconclusive, since Bn+1 would not be less for all n.

Answer 6

for \[\sum_{n=1}^{\infty}(-1)^na_n\]

all that you need is that \(a_n\) is decreasing for all n greater than some N.

Answer 7

so the A.S.T. will work here

Answer 8

I'm not sure how to show that the limit would 0 for it to converge... yet.

Answer 9

How about using the fact that \(\ln n<\sqrt n\) for all (positive integers) \(n\)? This comes from the fact that \(n<e^{n/2}\).

Then you have
\[\frac{\ln n}{n+3}<\frac{\sqrt n}{n+3}\sim\frac{1}{\sqrt n}\]

Answer 10

What is that squiggle saying?

Answer 11

It's a good idea SithsAndGiggles.

Answer 12

That comparison.

Answer 13

@SithsAndGiggles

Answer 14

The \(\sim\) is meant to communicate that \(\dfrac{\sqrt n}{n+3}\) behaves like \(\dfrac{1}{\sqrt n}\), since
\[\frac{\sqrt{n}}{n+3}\times\frac{\frac{1}{\sqrt n}}{\frac{1}{\sqrt n}}=\frac{1}{\sqrt n+\frac{3}{\sqrt n}}\]As \(n\to\infty\), you have that \(\dfrac{3}{\sqrt n}\to0\), and so \(\dfrac{1}{\sqrt n+\frac{3}{\sqrt n}}\) in turn behaves like \(\dfrac{1}{\sqrt n}\).

Answer 15

maybe just go a step further

\[\frac{\ln n}{n+3}<\frac{\sqrt n}{n+3}<\frac{\sqrt{n}}{n}=\frac{1}{\sqrt{n}}\]


though this doesn't show it is a decreasing function

Answer 16

Right, I assumed showing that \(b_n\to0\) was the OP's roadblock