Let X be a non-negative random variable, and $A_i$ be the event that $i-1 \leq X < i$.
(a) Show that $\sum_{i=1}^{\infty}(i-1)I_{Ai} \leq X < \sum_{i=1}^{\infty}iI_{A_i}$, where $I_{A_i}$ = 1 if $A_i$ occurs and = 0 otherwise.
Deduce directly that $\sum_{i=1}^{\infty}P(X \geq i) \leq \mathbf{E}[X] < 1 + \sum_{i=1}^{\infty}P(X \geq i )$
(b) By considering an exponential distribution for X of rate $\lambda$, deduce from part a that for $\lambda > 0$, $\lambda + e^{-\lambda} > 1$ and $(\lambda + 1)e^{-\lambda} \leq 1$.
Finally does the equation proved here have a name? Thank you.
a) If $j-1 \leq X < j$, $\sum_{i=1}^{\infty}(i-1)I_{Ai}= j-1$ and $\sum_{i=1}^{\infty}iI_{Ai} = j$. This is because only the $i=j$ term of the sums will contribute. So this proves your inequality. If now sum this inequality for every $X$ you get:
\begin{equation} \sum_{i=1}^{\infty}P(X \geq i) \leq \mathbf{E}[X] < 1 + \sum_{i=1}^{\infty}P(X \geq i ) \end{equation}
So I think that you made a typo. Finally for part b) you just have to substitute the mean and the cumulative of the exponential distribution in the formula above. You can notice that the two last inequalities can be proven also by Taylor expanding the exponential and noting that the rest is always positive.