I have the following problem:
Let $(X_n)$ be a sequence of i.i.d. exponential random variables with parameter $\lambda>0$. I need to show that $\forall \epsilon >0$ $$\sum_n \Bbb{P}\left(\frac{X_n}{log(n)}>\epsilon +\frac{1}{\lambda}\right)<\infty$$
My Idea was the following. Define $\Lambda_n=\left\{\frac{X_n}{log(n)}>\epsilon +\frac{1}{\lambda}\right\}=\left\{X_n\leq \log(n)\left(\epsilon +\frac{1}{\lambda}\right)\right\}^c$. Now I thought that for a fixed $n$ we have $$\Bbb{P}(\Lambda_n)=1-\Bbb{P}\left(\left\{X_n\leq \log(n)\left(\epsilon +\frac{1}{\lambda}\right)\right\}\right)=1-\int_{-\infty}^{\log(n)\left(\epsilon +\frac{1}{\lambda}\right)}\lambda e^{-\lambda x}dx$$
But somehow I get in trouble with this $-\infty$ since $\lambda >0$. So I wanted to ask, what did I oversee or is everything wrong?
Thanks for your help.
You can directly calculate.
You are actually making the mistake in the pdf of the exponential random variable.
If $X\sim \exp(\lambda) $ then the pdf $f(x)$ is given by:-
$f(x)=\begin{cases}\lambda e^{-\lambda x},\,x\geq 0\\0\,\text{elsewhere}\end{cases}$
$$P\left(\frac{X_n}{\ln(n)}>\epsilon +\frac{1}{\lambda}\right)=P(X_{n}>\ln(n)(\epsilon+\frac{1}{\lambda}))=\\\int_{\ln(n)(\epsilon+\frac{1}{\lambda})}^{\infty}\lambda e^{-\lambda x}\,dx=\exp\Big(-\lambda\ln(n)\Big(\epsilon+\frac{1}{\lambda}\Big)\Big)$$
$$=\exp\Big(\ln\Big(\frac{1}{n^{(\epsilon+\frac{1}{\lambda})}}\Big)\Big)$$
$$=\frac{1}{n^{\lambda(\epsilon+\frac{1}{\lambda})}}=\frac{1}{n^{\lambda\epsilon+1}}$$.
Now $\sum\frac{1}{n^{p}}$ converges if $p>1$ which it is in this case so you are done.