Series convergence for M-test

Question

How can show that $\sum_{n=0}^{\infty}ne^{-n \delta}$ is convergent where $\delta >0 $? I appreciate your kind help. Thank you!

Answer 1

$e^{n\delta}= $

$1+ n\delta +(n\delta)^2/2! + (n\delta)^3/3! +...\gt $

$(n\delta)^3/3!.$

$ne^{-n\delta}= \dfrac{n}{e^{n\delta}} \lt (\dfrac{3!}{\delta^3}) \dfrac {1}{n^2}.$

By comparison test $\sum ne^{-n\delta}$ converges.

Answer 2

Hint: Apply the integral test.

Answer 3

We have $e^{u}>u^{3}$ for large $u>0$, then $ne^{-n\delta}<n(n\delta)^{-3}=\delta^{-3}n^{-2}$ and $\displaystyle\sum n^{-2}<\infty$.

Answer 4

Another way: Note that $e^{-\delta}<1$ for $\delta>0$. Another interesting fact is that $$ \lim_{n\to \infty}n^{1/n}e^{-\delta}=e^{-\delta} $$