Cauchy's integral test for $\sum_{i=1}^\infty \frac{1}{n^n}$

Question

I was trying to check for limit of the series whose nth term is given below.

$$\sum_{i=1}^\infty \frac{1}{n^n}$$

Now $$\frac{T_n}{T_{n+1}} = (1+n){(1+\frac{1}{n})}^n$$ now this limit $\to \infty$ which means its reciprocal $\to 0$ and the whole thing converges but I was trying to prove it through the integral test but I am stuck at solving the integral part.

$$\int_1^\infty \frac{1}{x^x} dx$$

I took $y=x^x$ taking log both sides it becomes $\log y $ = $x$ $ \log x$ and differentiation turns out to be $$\frac{dy}{ydx} = 1+\log x$$ and that's where I need help. Will deeply appreciate some direction on this question.

Answer 1

I will suggest using comparison test. $$\sum_{n=1}^\infty\frac{1}{n^n}<\sum_{n=1}^\infty\frac{1}{n^2}<\infty$$

Answer 2

$\newcommand{d}[1]{\text{ d}#1}$ You don't need to evaluate for $\int_1^\infty \frac{1}{x^x} \d{x}$. Clearly $\int_1^2 \frac{1}{x^x} \d{x}$ is finite, and for $x \geq 2$ we have: $$ \int_2^\infty \frac{1}{x^x} \d{x} < \int_2^\infty \frac{1}{x^2} \d{x} < \infty $$ So the sum converges by the integral test.

Answer 3

As said, you face one of the Somophore's dreams.

Don't try anything and use numerical integration $$\int \frac{dx} {x^x}=0.70417$$

Answer 4

As others have noted, you don't need to evaluate the improper integral, you just need to show it converges. Here is one fun way to do so: Let $x=e^u$, so that

$$\int_1^\infty{dx\over x^x}=\int_0^\infty{e^u\,du\over e^{ue^u}}=\int_0^\infty{du\over e^{u(e^u-1)}}\lt\int_0^\infty {du\over e^{u^2}}\lt\int_0^\infty{du\over1+u^2}=\arctan u\big|_0^\infty={\pi\over2}$$

where we've made use of the inequality $e^v\gt1+v$ twice, first with $v=u$, then with $v=u^2$.