Question

Can anyone help me figure out if the series in the following link either diverges or converges via integral test or by asubn = 1/n, or some other method? Link>>>http://www.HostMath.com/Show.aspx?IsAsc=True&Code=%5Csum_%7Bk%3D1%7D%5E%5Cinfty%5Cke%5E%7B-3k%5E%7B2%7D%7D Thanks!

Answer 1

\[\sum_{k}^{\infty} k e^{-3k^2}\]You can use the direct comparison test to show that this series converges.

Since we know that the series 1/k^2 converges,\[\sum_{k}^{\infty} \frac {1}{k^2} = \frac {\pi^2}{6}\]And if we have the two functions \[f(k) = k e^{-3 k^2}, g(k) = \frac {1}{k^2}\] We know that f(k) < g(k) for all k > 1, so \[\sum_{k}^{\infty} f(k) \]must converge if \[\sum_{k}^{\infty} g(k)\] converges.

Answer 2

how did you automatically know to compare it it 1/k^2...is that just from experience or should I be looking at this problem from a different perspective?

Answer 3

Firs thing to really make it obvious for yourself would be to re-write the expression like this:\[\bf \sum_{k=1}^{\infty}\frac{ k }{ e^{3k^2} }\]Just by looking at this, we can tell that the series will eventually converge. To prove this, I feel that the easiest test would be the Ratio Test.\[\bf \lim_{k \rightarrow \infty} \left| \frac{ k +1}{ e^{3(k+1)^2} } \right| \div \left| \frac{ k }{ e^{3k^2} } \right|=\lim_{k \rightarrow \infty}e^{-6k-3}=0<1\] Clearly the series converges.

@SinginDaCalc2Blues

Answer 4

converges

Answer 5

Knowing to compare it to the p-series is sort of from experience. I haven't touched series in months, so using the direct comparison test was easier for me than using the ratio test. However, I think it would be more natural/obvious to use ratio test for this problem, as genius12 did.