If $X_i \sim \operatorname{Exponential}(\lambda)$ are iid such that $$f_{X_i}(x) = \lambda e^{-\lambda x}, \quad x > 0,$$ then $T = \sum_{i=1}^n X_i \sim \operatorname{Gamma}(n,\lambda)$ with $$f_T(x) = \frac{\lambda^n x^{n-1} e^{-\lambda x}}{\Gamma(n)}, \quad x > 0.$$ Then it is easy to see that $$\operatorname{E}[n/T] = \int_{x=0}^\infty \frac{n}{x} f_T(x) \, dx = \frac{n\lambda}{n-1} \int_{x=0}^\infty \frac{\lambda^{n-1} x^{n-2} e^{-\lambda x}}{\Gamma(n-1)} \, dx = \frac{n\lambda}{n-1},$$ for $n > 1$, since the last integral is simply the integral of of a $\operatorname{Gamma}(n-1,\lambda)$ PDF and is equal to $1$.
The distribution of $1/T$ is inverse gamma; i.e., $$T^{-1} \sim \operatorname{InvGamma}(n,\lambda)$$ with $$f_{1/T}(x) = \frac{\lambda^n e^{-\lambda/x}}{x^{n+1} \Gamma(n)}, \quad x > 0.$$
There is likely a proof somewhere on this site but I could not find it. Here I give a quick proof of my comment (since I originally mis-stated the result by forgetting the "lower bounded" restriction):
Let $\{X_i\}_{i=1}^{\infty}$ be a sequence of random variables, not necessarily identically distributed and not necessarily independent, that satisfy:
i) $E[X_i]=m_i$, where $m_i \in \mathbb{R}$ for all $i\in\{1, 2, 3, ...\}$.
ii) There is a constant $\sigma^2_{bound}$ such that $Var(X_i) \leq \sigma^2_{bound}$ for all $i \in \{1, 2, 3, ...\}$.
iii) The variables are pairwise uncorrelated, so $E[(X_i-m_i)(X_j-m_j)]=0$ for all $i \neq j$.
iv) There is a value $b \in \mathbb{R}$ such that, with prob 1, $X_i-m_i\geq b$ for all $i \in \{1, 2, 3, ...\}$.
Define $L_n = \frac{1}{n}\sum_{i=1}^n (X_i-m_i)$. Then $L_n\rightarrow 0$ with prob 1.
Proof: Since the variables are pairwise uncorrelated with bounded variance, we easily find for all $n$:
$$ E[L_n^2] = \frac{1}{n^2}\sum_{i=1}^n \sigma_i^2 \leq \frac{\sigma_{bound}^2}{n} $$
Fix $\epsilon>0$. It follows that:
$$ P[|L_n|>\epsilon] = P[L_n^2 \leq \epsilon^2] \leq \frac{E[L_n^2]}{\epsilon^2} \leq \frac{\sigma_{bound}^2}{n\epsilon^2} $$
Hence:
$$ \sum_{n=1}^{\infty} P[|L_{n^2}|>\epsilon] \leq \sum_{n=1}^{\infty}\frac{\sigma_{bound}^2}{n^2\epsilon^2} < \infty $$
and so $L_{n^2}\rightarrow 0$ with probability 1 by the Borel-Cantelli Lemma. That is, the $L_n$ values converge over the sparse subsequence $n\in\{1, 4, 9, 16, ...\}$.
Since $L_n \geq b$ for all $n$ and $L_{n^2}\rightarrow 0$ with probability 1, it can be shown that $L_n\rightarrow 0$ with probability 1. $\Box$
The lower bounded condition is typically treated by writing $X_n = X_n^+ - X_n^-$ where $X_n^+$ and $X_n^-$ are nonnegative and defined $X_n^+=\max[X_n,0]$, $X_n^-=-\min[X_n,0]$. If $X_n$ and $X_i$ are independent, then $X_n^+$ and $X_i^+$ are also independent. So the lower bounded condition can be removed for the case when variables are independent. However, if $X_n$ and $X_i$ are uncorrelated, that does not mean $X_n^+$ and $X_i^+$ are uncorrelated. So it is not clear to me if the lower-bounded condition can be removed when "independence" is replaced by the weaker condition "pairwise uncorrelated."
Best Answer
This is confusing limsup of events and limsup of random variables. Here, each $X_n$ is a random variable hence $$\limsup\limits_{n\to\infty}X_n/\log n$$ is the random variable $Y$ such that, for every $\omega$ in $\Omega$, $$ Y(\omega)=\limsup\limits_{n\to\infty}X_n(\omega)/\log n. $$ And the task is to prove that the event $[Y=1/\lambda]$ has probability $1$.
Note that the question most probably does not reproduce faithfully the text of the exercise asked. For example, $$ \text{lim(n→∞)supXn/logn} $$ should read $$ \limsup\limits_{n\to\infty}X_n/\log n, $$ where $\limsup\limits_{n\to\infty}$ acts as a single operation, or, equivalently, $$ \limsup\limits_{n\to\infty}\frac{X_n}{\log n}. $$