Question

Weierstrass M-test

Answer 1

With Weierstrass M-test I have to show that the following function range converges uniformly:
\[\sum_{n=0}^{\infty}n*e^{-n*x}\]\[on [1,\infty]\]

Answer 2

I now that \[\sum_{n=0}^{\infty}n*e^{-n*x}=\sum_{n=0}^{\infty}n*\frac{ 1 }{ e^{n*x} }\]
But is it okay to say that it converges because \[p=n*x>1\]

Answer 3

@jim_thompson5910

Answer 4

sorry I've never heard of this test before

Answer 5

Okay. But if you just luck at \[n*\frac{ 1 }{ e^{n*x} }\]would you say that it converges?

Answer 6

well I would argue that as n --> infinity, e^(nx) ---> infinity

so as n --> infinity, 1/e^(nx) ---> 0

Answer 7

so n*(1/e^(nx)) would turn into infinity * 0 which is an indeterminate form

Answer 8

See http://en.wikipedia.org/wiki/Weierstrass_M-test

you need to find another series that term by term is bigger than your series, but you can show converges

for example, you can show n/2^n > n/e^(nx)
and by the ratio test, show the series n/2^n converges.

Answer 9

Thank you.. @jim_thompson5910 & @phi

Answer 10

does this apply always \[\frac{ x }{ e^{x} }\le \frac{ x }{ 2^x }\] [1,∞]
???

Answer 11

for x≥1
e is about 2.7
so
\[ \frac{x}{2.7^x }≤ \frac{x}{2^x} \]
the denominator on the left is bigger, so we get a smaller number.

Answer 12

Thank you. It was very useful