Does This Series Converge for Different Values of Alpha?

In summary: In the first case, the series is convergent for all ##n## and in the second case, the series is divergent for all ##n##.
  • #1
DottZakapa
239
17
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
 
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  • #2
Mmm... is this series even defined for ##n=0##?
 
  • #3
Gaussian97 said:
Mmm... is this series even defined for ##n=0##?
sorry typing mistake is ##n=1##
 
  • #4
DottZakapa said:
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
DottZakapa said:
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##.
 

Related to Does This Series Converge for Different Values of Alpha?

1. What is the definition of the convergence of a series?

The convergence of a series refers to the behavior of the terms in the series as the number of terms increases. A series is said to converge if the terms approach a finite limit as the number of terms increases, and diverge if the terms do not approach a finite limit.

2. How is the convergence of a series determined?

The convergence of a series can be determined by using various tests such as the ratio test, the root test, or the comparison test. These tests compare the series to known convergent or divergent series to determine its behavior.

3. What is the significance of the convergence of a series?

The convergence of a series is important in mathematics as it allows us to determine the sum of an infinite series. It also has applications in physics, engineering, and other fields where infinite series are used to model real-life phenomena.

4. Can a series converge to more than one limit?

No, a series can only converge to one limit. This is because if the terms in a series are approaching two different limits, then the series cannot have a finite sum as it would be constantly switching between the two limits.

5. How does the convergence of a series relate to the convergence of its subsequences?

If a series converges, then all of its subsequences will also converge to the same limit. However, the converse is not always true - a series can have subsequences that converge to different limits or even diverge, even if the original series converges.

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