# Study the convergence and absolute convergence of the following series

• DottZakapa

#### DottZakapa

Homework Statement
study convergence and absolute convergence
Relevant Equations
numerical series
## \sum_{n=1}^\infty (-1)^n \frac {log(n)}{e^n}##

i take the absolute value and consider just

## \frac {log(n)}{e^n}##

i check by computing the limit if the necessary condition for convergence is satisfied

##\lim_{n \rightarrow +\infty} \frac {log(n)}{e^n} =\lim_{n \rightarrow +\infty} \frac {1}{ne^n}=0 ##

condition satisfied, now how do i find the rest? with which function can i compare it in order to find if it absolutely converges or not?
in the sense that, at this point i should find some serie for which i know the behaviour, then through comparison or asymptotic comparison i ca find out if the series converges or not.

Last edited:

Homework Statement:: study convergence and absolute convergence
Homework Equations:: numerical series

## \sum_{n=1}^\infty (-1)^n \frac {log(n)}{e^n}##

i take the absolute value and consider just

## \frac {log(n)}{e^n}##

i check by computing the limit if the necessary condition for convergence is satisfied

##\lim_{n \rightarrow +\infty} \frac {log(n)}{e^n} =\lim_{n \rightarrow +\infty} \frac {1}{ne^n}=0 ##

condition satisfied, now how do i find the rest?
But this condition is not sufficient. For example, the same limit for the series ##\sum_{n=1}^\infty \frac 1 n## is zero, but this series (harmonic series) is known to diverge.

Can you determine that the series itself converges - not the series of absolute values?

DottZakapa said:
with which function can i compare it in order to find if it absolutely converges or not?
in the sense that, at this point i should find some series for which i know the behaviour, then through comparison or asymptotic comparison i ca find out if the series converges or not.
What other tests for convergence do you know? Some common tests other than the comparison test are the limit comparison test, ratio test, and a couple of others.

BTW, "serie" isn't a word in English. "Series" is both singular and plural.

BTW, "serie" isn't a word in English. "Series" is both singular and plural.

Thanks for the correction :)

But this condition is not sufficient. For example, the same limit for the series ##\sum_{n=1}^\infty \frac 1 n## is zero, but this series (harmonic series) is known to diverge.

Yes is not sufficient but is a necessary according to one theorem.
But in case such limit is not zero, tells me that the series does not converges, hence i can stop there. If I understood correctly.

If it is zero i proceed applying the tests in order to find out if it converges or not.

Being a series with alternating signs, in order to study the simple and absolute convergence

1) I study the behaviour of ##\sum_{n=n0}^\infty |a_n|=\sum_{n=n0}^\infty b_n ##

2) If the series ##\sum_{n=n0}^\infty b_n## converges, then ##\sum_{n=n0}^\infty a_n ## converges absolutely and converges.

3) If the series ##\sum_{n=n0}^\infty b_n## diverges and ##\lim_{n \rightarrow +\infty} b_n =0 ## and the sequence ##{b_n}## is decreasing i can apply ##Leibniz## alternating series test.

So, now, in order to check point 2 i use one of the tests for positive series , which can be
- ##ratio~test## or
- ##root~test ## or
- ##comparison~test## or
-##asymptotic~comparison~test##

Am I correct?

Yes, looks good.

Homework Statement:: study convergence and absolute convergence
Homework Equations:: numerical series

## \sum_{n=1}^\infty (-1)^n \frac {log(n)}{e^n}##

i take the absolute value and consider just

## \frac {log(n)}{e^n}##
I proceeded as follows:

##\sum_{n=1}^\infty |(-1)^n \frac {log(n)}{e^n}| = \sum_{n=1}^\infty \frac {log(n)}{e^n}##

I apply the ## Root~Test##

##\lim_{n \rightarrow +\infty} \frac {log(n)}{e^n}~=~\lim_{n \rightarrow +\infty} \sqrt [n]{\frac {log(n)}{e^n}}~=~\lim_{n \rightarrow +\infty} \frac {log(n)^\left(\frac 1 n \right)}{e}~=~ \frac 1 e \lim_{n \rightarrow +\infty} log(n)^\left(\frac 1 n \right)##
the ##\lim_{n \rightarrow +\infty} \frac {log(n)} {n} =~0##, therefore, being ##l<1## the series converges absolutely.

Is the procedure correct?

Looks fine to me.