Determine whether the series is convergence or divergence?

[john458776]

\(\displaystyle \sum^\ \frac {k}{e^{k^2}}\)

So I used the Geometric series test to determine whether it converges or diverges.

\(\displaystyle \sum^\ k(\frac {1}{e^1})^{k^2}\)

[MATH]Since [/MATH]
[MATH]r < 1 ==> Converges[/MATH]
The series Converges.

I want to know whether I'm correct or the is something I did wrong. I'm still learning sequences and series.


Thank You.

 

[Steven G]

There is a very special form for a p-series and is your series in that form?

 

[john458776]

There is a very special form for a p-series and is your series in that form?

Yes I think it's in that form.
a_n[MATH]=[/MATH][MATH]a[/MATH][MATH]r[/MATH]^n-1

But the problem I'm trying to solve it's a_k[MATH]=[/MATH][MATH]a[/MATH][MATH]r[/MATH]^k2, where [MATH]r=1/[/MATH]e¹.

 

[john458776]

There is a very special form for a p-series and is your series in that form?

Sorry I edited my problem, I made a mistake by saying P-series test. I used the geometric test.
Thank you.

 

[Steven G]

So exactly what is a and r in your case? Is r being raised to the correct power?

 

[john458776]

So exactly what is a and r in your case? Is r being raised to the correct power?

a is the first term of the sequence, and r is the common ratio. I think r is not raised to the correct power. That's what confused me. I thought that maybe I should use the integral test not the geometric test because of that power.

 

[Steven G]

I did not ask for the meaning of a and r. I asked what is a and r in your problem?

Actually you did that r=1/e. Thank you. You claim that a is the value of the 1st term. Good! Now tell us what that value is. Is the value of the 2nd term as expected?

 

[Steven G]

I suspect that you are assuming that a=k. But k is changing! Your exponent is not increasing by 1 each time k changes.

Summation ar^n-1 = a + ar + ar² + ar³+ ar⁴+ ar⁵ + ...

While summation kr^k2 = 1r¹ + 2r⁴+ 3r⁹+ 4r¹⁶+..., where r=1/3.

Sorry, but they do not look similar to me.

 

[Steven G]

The integral test should work.

Lets suppose that the multiple k was a constant, say a.

Then summation a(1/e)^k2 would converge by the direct comparison test to the summation a(1/e)^k. Note that the series a(1/e)^k2 has just some of the terms of a(1/e)^k. That is, it was the multiple k that gave you more trouble then the exponent being k². I hope that you see this.

 

[pka]

\(\displaystyle \sum^\ \frac {k}{e^{k^2}}\)
So I used the Geometric series test to determine whether it converges or diverges.

Do Not over-think this question. That is all you have posted.
Apply the simple ratio test.

 

[john458776]

I suspect that you are assuming that a=k. But k is changing! Your exponent is not increasing by 1 each time k changes.

Summation ar^n-1 = a + ar + ar² + ar³+ ar⁴+ ar⁵ + ...

While summation kr^k2 = 1r¹ + 2r⁴+ 3r⁹+ 4r¹⁶+..., where r=1/3.

Sorry, but they do not look similar to me.

Thank you. Now I notice that k is not equal to a. Using the geometric test wont work in this particular problem. K is not constant it will keep on changing in each term.