Series converges and diverges with different tests.

[molten tusk]

How can the sum start at n = 0 if we divide by n?

[vestal mango]

ah sorry

it starts at n=1

just replace it everywhere

[molten tusk]

$\sum_{n=1}^{\infty} \frac{2^n+\big(-\frac{4}{3}\big)^n}{n}$

[quaint pagoda]

adonhs

[vestal mango]

and i wanted to aproximate it with $3^n > n$

[quaint pagoda]

Yanek

[vestal mango]

is that not correct what i have done above?

[molten tusk]

The sequence

[vestal mango]

what is wrong?

if the demoninator is bigger

the sequence should be smaller

[molten tusk]

What is $\lim_{n\to\infty} \frac{2^n+\big(-\frac{4}{3}\big)^n}{n}$?

[quaint pagoda]

adonhs

[vestal mango]

it will be $lim_{n \to \infty} 2^n ln 2 + (-4/3)^n \ln (-4/3) = \infty$

[quaint pagoda]

Yanek

[vestal mango]

ah no

im not sure

what to do about the (-4/3)^n

[molten tusk]

hmm

[vestal mango]

because ln x is only defined for x > 0

so it doesnt make sense here to take derrivative like this

and its sequence

so you can't use d'hopital rule

[molten tusk]

$\lim_{n\to\infty} \frac{2^n+(-1)^n\big(\frac{4}{3}\big)^n}{n} = \lim_{n\to\infty} \frac{2^n}{n}+(-1)^n\frac{\big(\frac{4}{3}\big)^n}{n}$

[vestal mango]

i don't know it looks like it diverges im not sure how to calculate this limit

because its infinity/infinity

and i cant use d'hopital as i said

[quaint pagoda]

adonhs

[vestal mango]

thats why i wanted to use different test

like the comprision test

[molten tusk]

well the comparison test failed because you have to compare up

$\sum_{n=1}^{\infty} \frac{2^n+\big(-\frac{4}{3}\big)^n}{n} = \sum_{n=1}^{\infty} (-1)^n\frac{2^n}{n}+\sum_{n=1}^{\infty}\frac{\big(\frac{4}{3}\big)^n}{n}$

[quaint pagoda]

adonhs

[molten tusk]

Now we can use Leibnitiz

$\sum_{n=1}^{\infty} (-1)^n\frac{2^n}{n}$ now do:

$\ \lim_{n\to\infty} \bigg|(-1)^n\frac{2^n}{n}\bigg| = \lim_{n\to\infty} \frac{2^n}{n} \neq 0 \Rightarrow :$ no convergence

[quaint pagoda]

adonhs

[molten tusk]

$\sum_{n=1}^{\infty}\frac{\big(\frac{4}{3}\big)^n}{n}$ check for zero sequence:

$\ \lim_{n\to\infty} \frac{\big(\frac{4}{3}\big)^n}{n} = \infty \neq 0 \Rightarrow :$ no convergence

[quaint pagoda]

adonhs

[molten tusk]

Your problem is also that you compare something that is convergent but it has to be a consistent comparison

Like yes

1/n >= 1/3^n

But Σ 1/n diverges and Σ 1/3^n does not

[vestal mango]

@molten tusk im not sure i follow you

[vestal mango]

also this is not complete prof

[vestal mango]

the only assumption that requires to use the comprision test is a_n and b_n > 0