# The convergence of a series

## 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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Mmm... is this series even defined for $n=0$?

Mmm... is this series even defined for $n=0$?
sorry typing mistake is $n=1$

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.

vela
Staff Emeritus
$$\sum_{n=0}^\infty \left( ne^{\frac 3 n}-n \right) \left ( \sin \frac {\alpha} {n} - \frac 5 n\right) \sim 3 \left ( \frac {\alpha} {n} - \frac 5 n\right)$$
You need to consider the cases where $\alpha=5$ and where $\alpha \ne 5$.