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

Mathematics Asked by Ramana on December 5, 2020

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.

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.

Correct answer by Eevee Trainer on December 5, 2020

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.

Answered by Noah Tang on December 5, 2020

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.

Answered by Davide Ravasini on December 5, 2020

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