Integral Test — Question 6

*Question 6 Consider the rapidly decaying series $$\sum_{n=1}^{\infty}n e^{-n^2}.$$

1. Verify that $f(x)=xe^{-x^2}$ satisfies the Integral Test hypotheses on $[1,\infty)$.

2. Evaluate the improper integral with a substitution and classify the series.

3. Derive upper and lower bounds for the remainder $R_N$.

4. Explain what the bound says about the speed of convergence.

*Question 6 – Solution

Let $f(x)=xe^{-x^2}$ for $x\ge1$. It is continuous and positive, and $$f'(x)=e^{-x^2}(1-2x^2)<0\qquad(x\ge1),$$ so it decreases.

With $u=x^2$ and $du=2x\,dx$, $$\int_1^{\infty}xe^{-x^2}\,dx =\frac12\int_1^{\infty}e^{-u}\,du=\frac1{2e}.$$ Thus the series converges by the Integral Test.

Since $\int_A^\infty xe^{-x^2}dx=\tfrac12e^{-A^2}$, $$\boxed{\frac12e^{-(N+1)^2}\le R_N\le\frac12e^{-N^2}}.$$ The Gaussian factor makes the tail decrease exceptionally rapidly.