Prove that the series $\sum \limits_{n=1}^{\infty}\int\limits_{n}^{n+1}e^{-x^2}dx$ converges I need to prove that the following series converges:
$$\sum_{n=1}^{\infty}\int\limits_{n}^{n+1}e^{-x^2}dx$$
I thought about approximating the integrand in order to use the limit comparison test, but using Taylor for the approximation will not help me, so I don't know what I can do.
 A: Although Fred's upper bound is nicer, you can also note that $\int_n^{n+1}e^{-x^2}\,dx < \int_n^{n+1} e^{-x}\,dx = e^{-n}-e^{-n-1}$, which means that the original series is bounded by a difference of convergent geometric series and is therefore convergent.
A: You can use the fundamental theorem of calculus to assume the existence of some function $F:[0,\infty)\to\mathbb{R}$ such that $F'(x)=e^{-x^2},$ and thus, $$\int_n^{n+1}e^{-x^2}\,\mathrm{d}x=F(n+1)-F(n).$$ Therefore, $$\sum_{n=1}^{\infty}\int_n^{n+1}e^{-x^2}\,\mathrm{d}x=\sum_{n=1}^{\infty}F(n+1)-F(n)=\lim_{N\to\infty}F(N+1)-F(1).$$ Therefore, the convergence of the series is equivalent to the convergence of $F.$
The convergence of $F$ can be investigated by choosing $F$ to be defined by $$F(x)=\int_0^xe^{-t^2}\mathrm{d}t.$$ So $$\lim_{N\to\infty}F(N+1)=\int_0^{\infty}e^{-t^2}\,\mathrm{d}t.$$ The convergence of this latter integral is a well-known result.
A: I think this may be the simplest method if you don't want to go too crazy with things.
Start by taking your sum on the left, and rewriting it as the latter integral.
$$
\sum_{n=1}^{\infty}\int_n^{n+1}e^{-x^2}\,\mathrm{d}x=\int_1^{\infty}e^{-x^2}\,\mathrm{d}x
$$
You are able to do this since the bounds of integration telescope from 1 to infinity.
Next simply noticing that $e^{-x^2}\le e^{-x}$, for all $x\ge1$, gives
$$\int_1^{\infty}e^{-x^2}\,\mathrm{d}x\le\int_1^{\infty}e^{-x}\,\mathrm{d}x$$
Then just solving and evaluating the last integral gives your original series to be strictly less than $\frac{1}{e}$, hence it is convergent.
If you're purely trying to solve this whilst keeping it in summation form, then you will have to use some other method than this one here.
