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Show that the following converges:

$$ \sum_{n=1}^{\infty }n^{3}e^{-t n^{2}+1} $$

where $t$ is a parameter.

  • 0
    what is $\alpha$ here? – 2017-01-26
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    Presumably $t>0$, or else is a complex number with positive real part? – 2017-01-26
  • 0
    t is real number – 2017-01-26
  • 0
    the exponential decreases faster then any polynomial can grow.... – 2017-01-26
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    Well the series (obviously) doesn't converge for $t\le 0$. – 2017-01-26

2 Answers 2

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If $t>0$ then $e^{-tn^2 +1} < n^{-5}$ for sufficiently large $n$ (it is true for any negative power), i.e. for $n>M$.

Now

$$\sum_{n=1}^{\infty} n^3e^{-tn^2 +1}=\sum_{n=1}^{M} n^3e^{-tn^2 +1}+\sum_{n=M+1}^{\infty} n^3e^{-tn^2 +1}$$

Note that it is enough to show that the second component is convergent . And by the choice of $M$:

$$\sum_{n=M+1}^{\infty} n^3e^{-tn^2 +1}<\sum_{n=M+1}^{\infty} n^3\cdot n^{-5}=\sum_{n=M+1}^{\infty}n^{-2}$$

It is well known that the last series is convergent as a piece of $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$. Thus the original series is convergent because $e^{-tn^2 +1} > 0$.

Note that for non-positive $t$ this is not true. For example for $t=-1$ obviously

$$\sum_{n=1}^{\infty} n^3e^{n^2+1}=\infty$$

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I would do it in two steps, first showing that $\sum n^3\exp(-tn)$ is convergent with the Ratio Test, and then comparing your series to this very favorably with a direct Comparison Test. Naturally, $t$ must be positive for all this to work.