Convergence of a series of integrals: $S=\sum_{n\geq 1}(-1)^{n}\int_{n}^{n+1}\frac{1}{t e^t}\,dt$ Let
$$S = -\int_{1}^{2}\frac{1}{te^t}\,dt + \int_{2}^{3}\frac{1}{te^t}\,dt-\int_{3}^{4}\frac{1}{te^t}\,dt + \cdots +\text{ad inf}$$
Does the series $S$ converge? Clearly,
$$S=\sum_{n\geq 1}(-1)^{n}\int_{n}^{n+1}\frac{1}{t e^t}\,dt$$
So I thought of using alternating series test as there is a $(-1)^{n}$ term, but I am not sure how to estimate the integral term.
 A: Notice that $1/te^t$ is a monotone decreasing function as $t$ grows. Thus (readily evident from the interpretation of an integral as signed area),
$$\int_{n}^{n+1} \frac{1}{te^t} dt \le \frac{1}{ne^n}$$
I believe this should be enough to get you to the desired conclusion.
A: The function
\begin{equation}
f(t)=\frac{1}{te^t}
\end{equation}
is continuous, strictly decreasing on $[1,+\infty)$ and its limit as $t\to+\infty$ is $0$. This is enough to conclude that the terms
\begin{equation}
a_n=\int_n^{n+1}\frac{1}{te^t}\,\text{d}t
\end{equation}
are decreasing and converge to $0$ as $n\to+\infty$. Indeed, by the mean value theorem, there are $t_n\in[n,n+1]$ and $t_{n+1}\in[n+1,n+2]$ such that $a_n=f(t_n)$ and $a_{n+1}=f(t_{n+1})$, thus $a_n=f(t_n)>f(t_{n+1})=a_{n+1}$ and $a_n=f(t_n)\to 0$ as $n\to+\infty$. All the requirements of the alternating series test are therefore fulfilled and allow to conclude that the series converges.
A: Set $$b_n=\int_{n}^{n+1}\frac{1}{t e^t}\,dt,$$
then $$\frac{1}{(n+1)e^{n+1}}\leq b_n\leq\frac{1}{ne^{n}},$$
and using squeeze theorem,
$$\lim_{n\to +\infty}{b_n}=0.$$
Obviously, $\{b_n\}$ is decreasing, therefore $S=\sum_n{(-1)^nb_n}$ is convergent.
