Convergence of a series of integrals: $S=sum_{ngeq 1}(-1)^{n}int_{n}^{n+1}frac{1}{t e^t},dt$

Question

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,
$$S=sum_{ngeq 1}(-1)^{n}int_{n}^{n+1}frac{1}{t e^t},dt$$
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.

Eevee Trainer · Accepted Answer

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.

Noah Tang · Answer

Set $$b_n=int_{n}^{n+1}frac{1}{t e^t},dt,$$
then $$frac{1}{(n+1)e^{n+1}}leq b_nleqfrac{1}{ne^{n}},$$
and using squeeze theorem,
$$lim_{nto +infty}{b_n}=0.$$
Obviously, ${b_n}$ is decreasing, therefore $S=sum_n{(-1)^nb_n}$ is convergent.

Davide Ravasini · Answer

The function
begin{equation}
f(t)=frac{1}{te^t}
end{equation}
is continuous, strictly decreasing on $[1,+infty)$ and its limit as $tto+infty$ 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 $nto+infty$. Indeed, by the mean value theorem, there are $t_nin[n,n+1]$ 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 $nto+infty$. All the requirements of the alternating series test are therefore fulfilled and allow to conclude that the series converges.

Convergence of a series of integrals: $S=sum_{ngeq 1}(-1)^{n}int_{n}^{n+1}frac{1}{t e^t},dt$

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