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Use Alternating series test for $ \sum_{n=0}^{\infty} (-1)^{n-1} n^2 e^{-(3+n^3)}$ . If the series converges whether it is absolutely converges or conditionally . (Hints. Integral test )

Answer: The n^th term is $ a_n=n^{2} e^{-(3+n^3)}$.

Now, $ \lim_{n \rightarrow \infty} a_{n} =0 $ . Also numeric value of each term is less than the preceding term.

Hence by Leibnitz test the series converges.

Now we have to check whether it converges absolutely or conditionally .

Consider the absolute series $ \sum_{n=0}^{\infty} n^2 e^{-(3+n^3)} $

**Here we have to apply integral test . For that let $ f(x)=x^2 e^{-(3+x^3)} $ , which is non-increasing and $ f(n)=a_n$.

Then , $ \lim_{M \rightarrow \infty } \int_{1}^{M} f(x) dx = ? $ . ** I need help here . **

** I can not apply. Any help is there ?**

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Since $f(n)=n^2 e^{-(3+n^3)}$ is continuous and strictly decreasing we can apply the integral test to find that $\sum f(n)$ converges if and only if $$I=\int_{0}^{\infty}x^2 e^{-(3+x^3)}dx$$ does. Applying the variable change $u=3+x^3$ and $du=3x^2$ we get $$\begin{aligned} I &= \frac{1}{3}\int_3^{\infty}e^{-u}du \\ &= \frac{1}{3}\left(\lim_{u\to\infty}(-e^{-u}) +e^{-3}\right) \end{aligned}$$ which you should be able to evaluate yourself, or at least determine the converge of.

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