Let $f(x)$ be the density function of the $X_i$, and $F(x)$ their cdf.
Let $Z$ be the maximum of the $X_i$. For completeness we find the distribution of $Z$, though that is likely familiar to you.
The event $Z\le z$ happens precisely if all the $X_i$ are $\le z$. This has probability $(F(z))^n$. Thus
$$F_Z(z)=(F(z))^n.$$
For the density function of $Z$, differentiate. We get $f_Z(z)=nf(z)(F(z))^{n-1}$.
Let $Y$ be the second largest of the $X_i$. We give a highly informal derivation of the density function $f_Y(y)$ of $Y$.
Let $dy$ be "small." We find the probability that $Y$ lies between $y$ and $y+dy$.
This will be approximately $f_Y(y)\,dy$.
Neglecting terms in higher powers of $dy$, the probability that the second largest lies between $y$ and $y+dy$ is the probability that some $X_i$ lies in this interval times the probability that $n-2$ of the $X_i$ lie below $y$ and $1$ lies above $y+dy$.
The $X_i$ that lies between $y$ and $y+dy$ can be chosen in $n$ ways. The probability it lies in the interval is approximately $F(y)\,dy$.
The probability that $n-2$ of the remaining $X_i$ lie below $y$, and $1$ lies above, is $\binom{n-1}{1}(1-F(y))^1 (F(y))^{n-2}$. "Thus,"
$$f_Y(y)=n\binom{n-1}{1}f(y)(1-F(y))(F(y))^{n-1}.$$
For the cdf $F_Y(y)$, integrate from $-\infty$ to $y$. The integral is in principle easy, make the substitution $u=F(y)$.
This can be proved by application of the 1st Borel-Cantelli lemma which states that: If $(A_n)_{n\in\mathbb{N}}$ is a (not necessarily independent) sequence of events such that $\sum_{n\in\mathbb{N}}\mathbb{P}(A_n)<\infty$, then $\mathbb{P}(A_n\ happens\ infinitely\ often)=0$. We will write $A_n\ i.o.$ to mean $A_n\ happens\ infinitely\ often$.
Now consider that
\begin{align}
limsup\frac{max\{X_1, ..., X_n\}}{n}\leq 1\ \ a.s. & \Leftrightarrow \mathbb{P}(limsup\frac{max\{X_1, ..., X_n\}}{n} > 1)=0 \\
& \Leftrightarrow\mathbb{P}(\forall m>N,\ sup_{n>m}\frac{max\{X_1, ..., X_n\}}{n}>1)=0 \\
& \Leftrightarrow\mathbb{P}(\frac{max\{X_1, ..., X_n\}}{n}>1\ \ \ i.o.)=0
\end{align}
This final implication can be seen from the fact that if $\frac{max\{X_1, ..., X_n\}}{n}>1$ does not happen for infinitely many $n$ (say that the largest $n$ for which this is true is $n'$), then $sup_{n>n'}\frac{max\{X_1, ..., X_n\}}{n}\leq 1$.
Hence applying the Borel-Cantelli lemma, we are left to show that $\sum_{n\in\mathbb{N}}\mathbb{P}(\frac{max\{X_1, ..., X_n\}}{n}>1)<\infty$.
We can calculate that:
\begin{align}
\mathbb{P}(\frac{max\{X_1, ..., X_n\}}{n}>1)&=\mathbb{P}(max\{X_1, ..., X_n\}>n)\\
&=\mathbb{P}(\{X_1>n\}\cup \{X_2>n\}\cup ... \cup \{X_n>n\})\\
&=\mathbb{P}(X_1>n)+\mathbb{P}(X_2>n)+...+\mathbb{P}(X_n>n)\\
&=n\mathbb{P}(X_1>n)
\end{align}
since the $X_i$s are i.i.d
And we see exactly why the conditions were given as they were, since the given condition $\sum_{n\in\mathbb{N}}n\mathbb{P}(X_1>n)<\infty$ implies that $\sum_{n\in\mathbb{N}}\mathbb{P}(\frac{max\{X_1, ..., X_n\}}{n}>1)<\infty$, and so we conclude, by the Borel-Cantelli lemma, that $limsup\frac{max\{X_1, ..., X_n\}}{n}\leq 1$ almost surely
Best Answer
Yes you can compute $\mathbb{P}(Y_n> y)$ as follows $$ \mathbb{P}(Y_n> y) = \mathbb{P}\left(\frac{n}{\max\{X_1, ..., X_n\}}> y\right) = \mathbb{P}\left({\max\{X_1, ..., X_n\}}< \frac{n}{y}\right) \\ =\mathbb{P}\left(X_1< \frac{n}{y}, ..., X_n< \frac{n}{y}\right) = \mathbb{P}\left(X_1< \frac{n}{y}\right)^n. $$ Therefore $$ \mathbb{P}(Y_n\leq y)=1-\mathbb{P}(Y_n> y) =1-\left(\frac{1}{\pi}\tan^{-1}{\frac{n}{y}}+\frac{1}{2}\right)^n. $$ Can you take the limit now?