(a)Show that the series$$\sum_{n=1}^\infty(\frac{e}{n})^n$$converges and use an appropriate convergence test to deduce that the improper integral $$\int_1^\infty\frac{e^y}{y^y}dy$$converges
(b)Using an appropriate convergence test and the fact that the improper integral in part (a) converges, show that $$\sum_{n=2}^\infty\frac{1}{(\ln(n))^{\ln(n)}}$$converges
(a)Use Ratio Test for $$\sum_{n=1}^\infty(\frac{e}{n})^n$$
That $$\left|\frac{a_n+1}{a_n}\right|
=\left|\frac{(\frac{e}{n+1})^{n+1}}{(\frac{e}{n})^n}\right|
=\left|\frac{(\frac{e^{n+1}}{(n+1)^{n+1}})}{(\frac{e^n}{n^n})}\right|
=\left|\frac{n^ne^{n+1}}{e^n(n+1)^{n+1}}\right|
=\left|\frac{n^ne}{(n+1)^{n+1}}\right|$$
Then take the limit we have
$$\lim_{n\rightarrow\infty}\left|\frac{n^ne}{(n+1)^{n+1}}\right|$$
Since $\forall x>0,\frac{n^ne}{(n+1)^{n+1}}>0$, implies
$$\lim_{n\rightarrow\infty}\left|\frac{n^ne}{(n+1)^{n+1}}\right|
=\lim_{n\rightarrow\infty}\frac{n^ne}{(n+1)^{n+1}}
=e\lim_{n\rightarrow\infty}\frac{n^n}{(n+1)^{n+1}}
=e\lim_{n\rightarrow\infty}\frac{n^n}{(n(1+\frac{1}{n}))^{n(1+\frac{1}{n})}}$$
$$=e\lim_{n\rightarrow\infty}\frac{n^n}{n^{n(1+\frac{1}{n})}(1+\frac{1}{n})^{n(1+\frac{1}{n})}}
=e\lim_{n\rightarrow\infty}\frac{1}{n^{n(1+\frac{1}{n})-n}(1+\frac{1}{n})^{n(1+\frac{1}{n})}}$$
$$=e\lim_{n\rightarrow\infty}\frac{1}{n(1+\frac{1}{n})^{n(1+\frac{1}{n})}}=0<1$$
Therefore by Ratio Test we have
$$\sum_{n=1}^\infty(\frac{e}{n})^n \text{ is convergent}$$
And the second part use Integral Test to show
$$\int_1^\infty\frac{e^y}{y^y}dy\text{ is convergent}$$
Let $f(x)=\frac{e^x}{x^x}$
Since $\forall x \in[1,\infty),f(x)$ is continuous positive and decreasing
We can conclude that $$\sum_{n=1}^\infty(\frac{e}{n})^n \text{ is convergent} \Leftrightarrow\int_1^\infty\frac{e^y}{y^y}dy\text{ is convergent}$$
By first part we can conclude that $$\int_1^\infty\frac{e^y}{y^y}dy\text{ is convergent}$$
(b) by (a) we have $$\int_1^\infty\frac{e^y}{y^y}dy\text{ is convergent}$$
Let $y=ln(x)$ that $dy=\frac{1}{x}dx$, therefore:
$$\int_1^\infty\frac{e^{\ln(x)}}{\ln(x)^{\ln(x)}}\frac{dx}{x}\text{ is convergent}$$
$$\Leftrightarrow \int_1^\infty\frac{1}{\ln(x)^{\ln(x)}}dx\text{ is convergent}$$
Since $\forall x \in[1,\infty),\frac{1}{\ln(x)^{\ln(x)}}$ is continuous positive and decreasing
By Integral Test we have:
$$\sum_{n=1}^\infty\frac{1}{\ln(x)^{\ln(x)}}dx\text{ is convergent}$$
And we know that:$$\sum_{n=1}^\infty\frac{1}{\ln(x)^{\ln(x)}}dx=\frac{1}{\ln(1)^{\ln(1)}}+\sum_{n=2}^\infty\frac{1}{\ln(x)^{\ln(x)}}dx$$
Which implies $$\sum_{n=2}^\infty\frac{1}{\ln(x)^{\ln(x)}}dx\text{ is convergent}$$
Best Answer
b): The series converges iff $\int_1^{\infty} \frac 1 {(\ln\,x )^{(\ln\,x )}} dx <\ \infty$. In this put $y=\ln\, x$. You will see that this transforms to $\int \frac {e^{y}} {y^{y}} dy$.