Nothing like little Sunday morning series action! x s.t. sum{0,inf}x^(n)/n! converges

[frank789]

Hi all!

I have a question, the likes of which I have never done before and cannot find a solution on the internet that satisfied me...hence I turn to the professionals.

I am asked to find the value of x such that sum{0,inf}x^(n)/n! converges and state the interval and radius of convergence.

I wrote out my series, {x/1},{x^2/2!}.....

then I did comparison test and simplified down to (x/(n+1))

I argued all x where x<(n+1) would make the series convergent since it would give a ratio of less than 1 with comparison test. I was a little fuzzy after this. Would if be fair to assume x gets replaced with n? If that is the case I can confidently say the the interval of convergence would be (-inf, inf) with the radius being inf. That is the posted answer for question but is there another path I should be taking to arrive at this destination?

Thanks

 

[frank789]

ok just tried another one

sum{1,inf}(x^n)(n^n)

used root test and got xn

i know the series converges if xn < 1 so I wrote x<n but im stuck!

 

[Deleted member 4993]

Hi all!

I have a question, the likes of which I have never done before and cannot find a solution on the internet that satisfied me...hence I turn to the professionals.

I am asked to find the value of x such that sum{0,inf}x^(n)/n! converges and state the interval and radius of convergence.

I wrote out my series, {x/1},{x^2/2!}.....

then I did comparison test and simplified down to (x/(n+1))

I argued all x where x<(n+1) would make the series convergent since it would give a ratio of less than 1 with comparison test. I was a little fuzzy after this. Would if be fair to assume x gets replaced with n? If that is the case I can confidently say the the interval of convergence would be (-inf, inf) with the radius being inf. That is the posted answer for question but is there another path I should be taking to arrive at this destination?

Thanks

Look at the Taylor series expansion of e^x around x=0

 

[frank789]

ok so I missed the first term

I should have the first term as x^0/0! which is 1 which does not contain x

so that is why x does not matter?

 

[Steven G]

ok just tried another one

sum{1,inf}(x^n)(n^n)

used root test and got xn

i know the series converges if xn < 1 so I wrote x<n but im stuck!

Why does xn<1 imply x<n?
If you multiply two numbers and the result is less than 1, it is probable that one of those numbers is less than the other but why does the number that we happen to call x have to be small than the number we happen to call n? I don't get that one.
For example, if x=3/4 and n=1/2, then xn=3/8<1, yet x>n. For the record, both numbers do not have to less than 1. For example x can be 4 and n can be 1/8. Their product is less than 1 while x>n. For x=.2 and n=.2, we have xn<1 while x=n.