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The convergence of a series

  • Thread starter DottZakapa
  • Start date
  • #1
145
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Homework Statement:

the series ##\sum_{n=0}^\infty \left( ne^{\frac 3 n}-n \right) \left ( \sin \frac {\alpha} {n} - \frac 5 n\right)##

Relevant Equations:

convergence of a series
The series ##\sum_{n=0}^\infty \left( ne^{\frac 3 n}-n \right) \left ( \sin \frac {\alpha} {n} - \frac 5 n\right)##

i did

##\sum_{n=0}^\infty n\left( e^{\frac 3 n}-1 \right) \left ( \sin \frac {\alpha} {n} - \frac 5 n\right)##

for n going to infinity

## \left( e^{\frac 3 n}-1 \right)## is asymptotic to ## \frac 3 n ## ##\Rightarrow## ##n \frac 3 n \Rightarrow 3##

## \left ( \sin \frac {\alpha} {n}\right)## is asymptotic to ##\frac {\alpha} {n}##

end i end up with
##\sum_{n=0}^\infty \left( ne^{\frac 3 n}-n \right) \left ( \sin \frac {\alpha} {n} - \frac 5 n\right)## ~ ##3 \left ( \frac {\alpha} {n} - \frac 5 n\right)##

is it correct up to here? can't see how to get a conclusion, supposing that is correct up to here.
an help?
don't understand why i struggle so much with series
 

Answers and Replies

  • #2
175
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Mmm... is this series even defined for ##n=0##?
 
  • #3
145
7
Mmm... is this series even defined for ##n=0##?
sorry typing mistake is ##n=1##
 
  • #4
WWGD
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Homework Statement:: the series ##\sum_{n=0}^\infty \left( ne^{\frac 3 n}-n \right) \left ( \sin \frac {\alpha} {n} - \frac 5 n\right)##
Relevant Equations:: convergence of a series

The series ##\sum_{n=0}^\infty \left( ne^{\frac 3 n}-n \right) \left ( \sin \frac {\alpha} {n} - \frac 5 n\right)##

i did

##\sum_{n=0}^\infty n\left( e^{\frac 3 n}-1 \right) \left ( \sin \frac {\alpha} {n} - \frac 5 n\right)##

for n going to infinity

## \left( e^{\frac 3 n}-1 \right)## is asymptotic to ## \frac 3 n ## ##\Rightarrow## ##n \frac 3 n \Rightarrow 3##

## \left ( \sin \frac {\alpha} {n}\right)## is asymptotic to ##\frac {\alpha} {n}##

end i end up with
##\sum_{n=0}^\infty \left( ne^{\frac 3 n}-n \right) \left ( \sin \frac {\alpha} {n} - \frac 5 n\right)## ~ ##3 \left ( \frac {\alpha} {n} - \frac 5 n\right)##

is it correct up to here? can't see how to get a conclusion, supposing that is correct up to here.
an help?
don't understand why i struggle so much with series
Series are not a simple topic.
But don't leave the ## \Sigma ## out :
## \Sigma 3 \left ( \frac {\alpha} {n} - \frac 5 n\right)=(3/n)( \alpha -5) ##

Now you can maybe do a L'Hopital-like thing and see if the n-th term goes to 0 fast-enough for convergence.
 
  • #5
vela
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I end up with
$$\sum_{n=0}^\infty \left( ne^{\frac 3 n}-n \right) \left ( \sin \frac {\alpha} {n} - \frac 5 n\right) \sim 3 \left ( \frac {\alpha} {n} - \frac 5 n\right)$$
Is it correct up to here? I can't see how to get a conclusion, supposing that is correct up to here. Any help? I don't understand why I struggle so much with series.
You need to consider the cases where ##\alpha=5## and where ##\alpha \ne 5##.
 

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