- #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
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