- #1

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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

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