Question

Prove that this is absolutely convergent

Answer 1

\[\sum_{n=1}^{\infty} \frac{ n+1 }{ n }a _{n}\] if \[\sum_{n=1}^{\infty} a _{n}\] is absolutely convergent

Answer 2

\[ \sum_{n=1}^{\infty} \frac{ n+1 }{ n }a _{n} = \sum_{n=1}^{\infty} \left(1+\frac{ 1 }{ n }\right)a _{n} = \sum_{n=1}^{\infty} a _{n} + \sum_{n=1}^{\infty} \frac{ 1 }{ n }a _{n} \]

Answer 3

yeah..i got that too but 1/n is a harmonic series right? so it diverge but how to prove it absolutely converge

Answer 4

try root test

Answer 5

since \(\sum \limits_{n=1}^{\infty} a _{n} \) is convergent, \(\lim \limits_{n\to \infty } |a_n|^{\frac{1}{n}} = L_0\)

Answer 6

use that result^ when applying root test for the series in question

Answer 7

i dont get how 1/n become the exponent suddenly

Answer 8

take \[\lim \limits_{n\to \infty } |\dfrac{1}{n}a_n|^{\frac{1}{n}} \],

and show that it evaluates the same limit value

Answer 9

but how do i do the root test when an is unknown?

Answer 10

1/n does not converge,
but \(a_n/n\) converges cuz \(a_n\) converges

Answer 11

use the previous \(L_0\) result

Answer 12

okay i think i get it, the resulting will be an/n after root test but, what rule is applicable to say an/n is convergent?

Answer 13

Here it is :

\[\lim \limits_{n\to \infty } |\dfrac{1}{n}a_n|^{\frac{1}{n}} = \lim \limits_{n\to \infty } |(\dfrac{1}{n})^{\frac{1}{n}}a_n^{\frac{1}{n}}| = \lim \limits_{n\to \infty } |(1)a_n^{\frac{1}{n}}| = \lim \limits_{n\to \infty } |a_n^{\frac{1}{n}}| = L_0 \]

Answer 14

\(L_0 < 1\) from the hypothesis
QED

Answer 15

eh uhm an^1/n isnt it an^0? so it is 1?

Answer 16

that result was GIVEN to us. we're not evaluating it.

Answer 17

we're given that \(\sum \limits_{n=1}^{\infty} a _{n}\) converges,


So,


\(\sum \limits_{n=1}^{\infty} a _{n}\) converges \( \iff\) \(\lim \limits_{n\to \infty } |a_n|^{\frac{1}{n}} = L_0 \)

Answer 18

http://prntscr.com/3dk099

Answer 19

haha okay xD, thanks! i think i get it now haha

Answer 20

np :)

earlier you're saying the exponent 1/n becomes 0, so the limit becomes 1.

take a look at :
http://www.wolframalpha.com/input/?i=limit+n+-%3E+%5Cinfty+n%5E%281%2Fn%29

Answer 21

\(\lim \limits_{n\to \infty } (\#\#)^{\frac{1}{n}} \) need not be 1 always

Answer 22

yeah, because if that was the case its inconclusive thats why im confused just now, thank you! it clear now :D

Answer 23

btw, i never did series before... so be careful when u dealing wid me lol.. .i might be wrong most of the time :P