Is this series convergent or divergent.

[nothingkwt]

Homework Statement

Ʃ ne^(-n2)

Homework Equations

The Attempt at a Solution

I used the ratio test and wanted to know if the way I did it is correct or not

|a_(n+1) / a_(n)|

n+1 (e^{(-n2 -2n-1)}) / n e^(-n2)

Now e^{-n^2} cancels and we get

lim_n→∞ n+1/n * 1/(e²ⁿ)(e)

After you take the limits you get (1)*0 = 0 < 1 so it's convergent

 

[pasmith]

Homework Statement

Ʃ ne^(-n2)

Homework Equations

The Attempt at a Solution

I used the ratio test and wanted to know if the way I did it is correct or not

|a_(n+1) / a_(n)|

(n+1) (e^{(-n2 -2n-1)}) / (n e^(-n2))

Now e^{-n^2} cancels and we get

lim_n→∞ (n+1)/n * 1/((e²ⁿ)(e))

After you take the limits you get (1)*0 = 0 < 1 so it's convergent

I have added your missing brackets. Aside from that your working is correct.

 

[Curious3141]

Homework Statement

Ʃ ne^(-n2)

Homework Equations

The Attempt at a Solution

I used the ratio test and wanted to know if the way I did it is correct or not

|a_(n+1) / a_(n)|

n+1 (e^{(-n2 -2n-1)}) / n e^(-n2)

Now e^{-n^2} cancels and we get

lim_n→∞ n+1/n * 1/(e²ⁿ)(e)

After you take the limits you get (1)*0 = 0 < 1 so it's convergent

Sure, this works. But this sum looks almost purpose-built for the integral test. Try it (if you're allowed to use it).

 

[nothingkwt]

Sure, this works. But this sum looks almost purpose-built for the integral test. Try it (if you're allowed to use it).

I am actually but I just wanted to see if the ratio test worked

Thanks for the replies!

 

[Ray Vickson]

Homework Statement

Ʃ ne^(-n2)

Homework Equations

The Attempt at a Solution

I used the ratio test and wanted to know if the way I did it is correct or not

|a_(n+1) / a_(n)|

n+1 (e^{(-n2 -2n-1)}) / n e^(-n2)

Now e^{-n^2} cancels and we get

lim_n→∞ n+1/n * 1/(e²ⁿ)(e)

After you take the limits you get (1)*0 = 0 < 1 so it's convergent

You really need to learn how to write in ASCII---in particular, you need to use brackets. Your first equation AS WRITTEN says
\left|\frac{a_{n+1}}{a_n}\right| = n +1 \frac{e^{-n^2-2n-1}}{n} e^{-n^2}.
Your second formula reads as
\lim_{n \to \infty} n + \frac{1}{n} \frac{1}{e^{2n}} e
I hope these are not what you mean. I hope you intended the first one to be
\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)e^{-n^2-2n-1}}{n e^{-n^2}},
etc. To make sure this happens you need parentheses!

 

[nothingkwt]

You really need to learn how to write in ASCII---in particular, you need to use brackets. Your first equation AS WRITTEN says
\left|\frac{a_{n+1}}{a_n}\right| = n +1 \frac{e^{-n^2-2n-1}}{n} e^{-n^2}.
Your second formula reads as
\lim_{n \to \infty} n + \frac{1}{n} \frac{1}{e^{2n}} e
I hope these are not what you mean. I hope you intended the first one to be
\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)e^{-n^2-2n-1}}{n e^{-n^2}},
etc. To make sure this happens you need parentheses!

Yeah I'm not very good with ASCII I'm still learning how to use them.